Finitely generated and residually finite implies Hopfian - Revision history

Vipul: /* Statement */

2012-01-23T00:56:42Z

Statement

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	==Statement==		==Statement==

	Any [[finitely generated residually finite group]] (i.e., a group that is both [[fact about::finitely generated group;2\| ]][[finitely generated group\|finitely generated]] and [[~~defining ingredient~~::residually finite group;1\| ]][[residually finite group\|residually finite]]) is a [[Hopfian group]].		Any [[finitely generated residually finite group]] (i.e., a group that is both [[fact about::finitely generated group;2\| ]][[finitely generated group\|finitely generated]] and [[fact about::residually finite group;1\| ]][[residually finite group\|residually finite]]) is a [[Hopfian group]].

	==Definitions used==		==Definitions used==

Vipul at 00:55, 23 January 2012

2012-01-23T00:55:46Z

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	{{group property implication\|		{{group property implication\|
	stronger = finitely generated residually finite group\|		stronger = finitely generated residually finite group\|
	weaker = Hopfian group}}		stronger relevance = 1\|
			weaker = Hopfian group\|
			weaker relevance = 1}}

	[[difficulty level::3\| ]]		[[difficulty level::3\| ]]

Vipul: /* Statement */

2012-01-23T00:55:14Z

Statement

← Older revision		Revision as of 00:55, 23 January 2012
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	==Statement==		==Statement==

	Any [[finitely generated residually finite group]] (i.e., a group that is both [[~~defining ingredient~~::finitely generated group\|finitely generated]] and [[defining ingredient::residually finite group\|residually finite]]) is a [[Hopfian group]].		Any [[finitely generated residually finite group]] (i.e., a group that is both [[fact about::finitely generated group;2\| ]][[finitely generated group\|finitely generated]] and [[defining ingredient::residually finite group;1\| ]][[residually finite group\|residually finite]]) is a [[Hopfian group]].

	==Definitions used==		==Definitions used==

Vipul at 16:57, 24 February 2011

2011-02-24T16:57:09Z

← Older revision		Revision as of 16:57, 24 February 2011
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	weaker = Hopfian group}}		weaker = Hopfian group}}

			[[difficulty level::3\| ]]
	==Statement==		==Statement==

Vipul: /* Proof */

2011-02-22T22:49:13Z

Proof

← Older revision		Revision as of 22:49, 22 February 2011
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	\| 2 \|\| All the homomorphisms <math>\alpha \circ \varphi^n</math>, for <math>n</math> varying over positive integers, are ''pairwise distinct'' homomorphisms from <math>G</math> to <math>K</math> \|\|<math>\varphi</math> is surjective, <math>g</math> is in the kernel of <math>\varphi</math> \|\| Step (1) \|\| <toggledisplay>Suppose <math>m < n</math>. We need to show that <math>\alpha \circ \varphi^m \ne \alpha \circ \varphi^n</math>. Since <math>\varphi</math> is surjective, <math>\varphi^m</math> is also surjective. In particular, there is <math>h \in G</math> such that <math>\varphi^m(h) = g</math>. Since <math>n > m</math> and <math>g</math> is in the kernel of <math>\varphi</math>, <math>\varphi^n(g)</math> is the identity element. Thus, <math>\alpha \circ \varphi^m</math> sends <math>h</math> to a non-identity element of <math>K</math> but <math>\alpha \circ \varphi^n</math> sends <math>h</math> to the identity element of <math>K</math>. Hence, these two maps are unequal.</toggledisplay>		\| 2 \|\| All the homomorphisms <math>\alpha \circ \varphi^n</math>, for <math>n</math> varying over positive integers, are ''pairwise distinct'' homomorphisms from <math>G</math> to <math>K</math> \|\|<math>\varphi</math> is surjective, <math>g</math> is in the kernel of <math>\varphi</math> \|\| Step (1) \|\| <toggledisplay>Suppose <math>m < n</math>. We need to show that <math>\alpha \circ \varphi^m \ne \alpha \circ \varphi^n</math>. Since <math>\varphi</math> is surjective, <math>\varphi^m</math> is also surjective. In particular, there is <math>h \in G</math> such that <math>\varphi^m(h) = g</math>. Since <math>n > m</math> and <math>g</math> is in the kernel of <math>\varphi</math>, <math>\varphi^n(g)</math> is the identity element. Thus, <math>\alpha \circ \varphi^m</math> sends <math>h</math> to a non-identity element of <math>K</math> but <math>\alpha \circ \varphi^n</math> sends <math>h</math> to the identity element of <math>K</math>. Hence, these two maps are unequal.</toggledisplay>
	\|-		\|-
	\| 3 \|\| There are only finitely many homomorphisms from <math>G</math> to <math>K</math> \|\| <math>G</math> is finitely generated \|\| \|\| <toggledisplay>A homomorphism of groups is completely specified by where the generators go. Since <math>G</math> has a finite generating set and <math>K</math> is finite, there are only finitely many possibilities for a homomorphism, bounded by <math>\|K\|^s</math> where <math>s</math> is the minimum size of generating set for <math>G</math>.</toggledisplay>		\| 3 \|\| There are only finitely many homomorphisms from <math>G</math> to <math>K</math> \|\| <math>G</math> is finitely generated \|\| Step (1) (specifically, that <math>K</math> is finite) \|\| <toggledisplay>A homomorphism of groups is completely specified by where the generators go. Since <math>G</math> has a finite generating set and <math>K</math> is finite, there are only finitely many possibilities for a homomorphism, bounded by <math>\|K\|^s</math> where <math>s</math> is the minimum size of generating set for <math>G</math>.</toggledisplay>
	\|-		\|-
	\| 4 \|\| We have the required contradiction \|\| \|\| Steps (2), (3) \|\| <toggledisplay>Step (2) gives infinitely many pairwise distinct homomorphisms from <math>G</math> to <math>K</math>, Step (3) asserts that there are only finitely many.		\| 4 \|\| We have the required contradiction \|\| \|\| Steps (2), (3) \|\| <toggledisplay>Step (2) gives infinitely many pairwise distinct homomorphisms from <math>G</math> to <math>K</math>, Step (3) asserts that there are only finitely many.

Vipul: /* Hands-on proof */

2011-02-22T22:42:54Z

Hands-on proof

← Older revision		Revision as of 22:42, 22 February 2011
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	\| 2 \|\| All the homomorphisms <math>\alpha \circ \varphi^n</math>, for <math>n</math> varying over positive integers, are ''pairwise distinct'' homomorphisms from <math>G</math> to <math>K</math> \|\|<math>\varphi</math> is surjective, <math>g</math> is in the kernel of <math>\varphi</math> \|\| Step (1) \|\| <toggledisplay>Suppose <math>m < n</math>. We need to show that <math>\alpha \circ \varphi^m \ne \alpha \circ \varphi^n</math>. Since <math>\varphi</math> is surjective, <math>\varphi^m</math> is also surjective. In particular, there is <math>h \in G</math> such that <math>\varphi^m(h) = g</math>. Since <math>n > m</math> and <math>g</math> is in the kernel of <math>\varphi</math>, <math>\varphi^n(g)</math> is the identity element. Thus, <math>\alpha \circ \varphi^m</math> sends <math>h</math> to a non-identity element of <math>K</math> but <math>\alpha \circ \varphi^n</math> sends <math>h</math> to the identity element of <math>K</math>. Hence, these two maps are unequal.</toggledisplay>		\| 2 \|\| All the homomorphisms <math>\alpha \circ \varphi^n</math>, for <math>n</math> varying over positive integers, are ''pairwise distinct'' homomorphisms from <math>G</math> to <math>K</math> \|\|<math>\varphi</math> is surjective, <math>g</math> is in the kernel of <math>\varphi</math> \|\| Step (1) \|\| <toggledisplay>Suppose <math>m < n</math>. We need to show that <math>\alpha \circ \varphi^m \ne \alpha \circ \varphi^n</math>. Since <math>\varphi</math> is surjective, <math>\varphi^m</math> is also surjective. In particular, there is <math>h \in G</math> such that <math>\varphi^m(h) = g</math>. Since <math>n > m</math> and <math>g</math> is in the kernel of <math>\varphi</math>, <math>\varphi^n(g)</math> is the identity element. Thus, <math>\alpha \circ \varphi^m</math> sends <math>h</math> to a non-identity element of <math>K</math> but <math>\alpha \circ \varphi^n</math> sends <math>h</math> to the identity element of <math>K</math>. Hence, these two maps are unequal.</toggledisplay>
	\|-		\|-
	\| 3 \|\| There are only finitely many homomorphisms from <math>G</math> to <math>K</math> \|\| \|\| \|\| <toggledisplay>A homomorphism of groups is completely specified by where the generators go. Since <math>G</math> has a finite generating set and <math>K</math> is finite, there are only finitely many possibilities for a homomorphism, bounded by <math>\|K\|^s</math> where <math>s</math> is the minimum size of generating set for <math>G</math>.</toggledisplay>		\| 3 \|\| There are only finitely many homomorphisms from <math>G</math> to <math>K</math> \|\| <math>G</math> is finitely generated \|\| \|\| <toggledisplay>A homomorphism of groups is completely specified by where the generators go. Since <math>G</math> has a finite generating set and <math>K</math> is finite, there are only finitely many possibilities for a homomorphism, bounded by <math>\|K\|^s</math> where <math>s</math> is the minimum size of generating set for <math>G</math>.</toggledisplay>
	\|-		\|-
	\| 4 \|\| We have the required contradiction \|\| \|\| Steps (2), (3) \|\| <toggledisplay>Step (2) gives infinitely many pairwise distinct homomorphisms from <math>G</math> to <math>K</math>, Step (3) asserts that there are only finitely many.		\| 4 \|\| We have the required contradiction \|\| \|\| Steps (2), (3) \|\| <toggledisplay>Step (2) gives infinitely many pairwise distinct homomorphisms from <math>G</math> to <math>K</math>, Step (3) asserts that there are only finitely many.

Vipul: /* Proof */

2011-02-22T22:42:24Z

Proof

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	===Hands-on proof===		===Hands-on proof===

			{{tabular proof format}}

	'''Given''': A finitely generated group <math>G</math> that is also residually finite. A surjective homomorphism <math>\varphi:G \to G</math>.		'''Given''': A finitely generated group <math>G</math> that is also residually finite. A surjective homomorphism <math>\varphi:G \to G</math>.
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	'''To prove''': <math>\varphi</math> is an automorphism of <math>G</math>.		'''To prove''': <math>\varphi</math> is an automorphism of <math>G</math>.

	'''Proof''': We prove this by contradiction. Suppose <math>\varphi</math> is not an automorphism of <math>G</math>. Then, since it is surjective, there exists an non-identity element <math>g</math> in its kernel.		'''Proof''': We prove this by contradiction.

			'''ASSUMPTION THAT WILL LEAD TO CONTRADICTION''': Suppose <math>\varphi</math> is not an automorphism of <math>G</math>. Then, since it is surjective, it must fail to be injective, so there exists an non-identity element <math>g</math> in its kernel.

	# There is a surjective homomorphism <math>\alpha:G \to K</math> for some finite group <math>K</math> such that <math>\alpha(g)</math> is not the identity element: <toggledisplay>Since <math>G</math> is residually finite, and <math>g</math> is a non-identity element, there is a normal subgroup <math>N</math> of finite index in <math>G</math> such that <math>g \notin N</math>. Let <math>K = G/N</math> and <math>\alpha:G \to K</math> be the quotient map.</toggledisplay>		{\| class="sortable" border="1"
	# All the homomorphisms <math>\alpha \circ \varphi^n</math> are distinct homomorphisms from <math>G</math> to <math>K</math>: <toggledisplay>Suppose <math>m < n</math>. We need to show that <math>\alpha \circ \varphi^m \ne \alpha \circ \varphi^n</math>. Since <math>\varphi</math> is surjective, <math>\varphi^m</math> is also surjective. In particular, there is <math>h \in G</math> such that <math>\varphi^m(h) = g</math>. Since <math>n > m</math> and <math>g</math> is in the kernel of <math>\varphi</math>, <math>\varphi^n(g)</math> is the identity element. Thus, <math>\alpha \circ \varphi^m</math> sends <math>h</math> to a non-identity element of <math>K</math> but <math>\alpha \circ \varphi^n</math> sends <math>h</math> to the identity element of <math>K</math>. Hence, these two maps are unequal.</toggledisplay>		! Step no. !! Assertion/construction !! Given data/assumptions used !! Previous steps used !! Explanation
	# There are only finitely many homomorphisms from <math>G</math> to <math>K</math> ~~: This follows from fact (1).~~ <toggledisplay>A homomorphism of groups is completely specified by where the generators go. Since <math>G</math> has a finite generating set and <math>K</math> is finite, there are only finitely many possibilities for a homomorphism, bounded by <math>\|K\|^s</math> where <math>s</math> is the minimum size of generating set for <math>G</math>.</toggledisplay>		\|-
	# We have the required contradiction ~~from~~ (2) ~~and~~ (3).		\| 1 \|\| There is a surjective homomorphism <math>\alpha:G \to K</math> for some finite group <math>K</math> such that <math>\alpha(g)</math> is not the identity element \|\| <math>G</math> is residually finite, <math>g</math> is a non-identity element. \|\| \|\| <toggledisplay>Since <math>G</math> is residually finite, and <math>g</math> is a non-identity element, there is a normal subgroup <math>N</math> of finite index in <math>G</math> such that <math>g \notin N</math>. Let <math>K = G/N</math> and <math>\alpha:G \to K</math> be the quotient map.</toggledisplay>
			\|-
			\| 2 \|\| All the homomorphisms <math>\alpha \circ \varphi^n</math>, for <math>n</math> varying over positive integers, are ''pairwise distinct'' homomorphisms from <math>G</math> to <math>K</math> \|\|<math>\varphi</math> is surjective, <math>g</math> is in the kernel of <math>\varphi</math> \|\| Step (1) \|\| <toggledisplay>Suppose <math>m < n</math>. We need to show that <math>\alpha \circ \varphi^m \ne \alpha \circ \varphi^n</math>. Since <math>\varphi</math> is surjective, <math>\varphi^m</math> is also surjective. In particular, there is <math>h \in G</math> such that <math>\varphi^m(h) = g</math>. Since <math>n > m</math> and <math>g</math> is in the kernel of <math>\varphi</math>, <math>\varphi^n(g)</math> is the identity element. Thus, <math>\alpha \circ \varphi^m</math> sends <math>h</math> to a non-identity element of <math>K</math> but <math>\alpha \circ \varphi^n</math> sends <math>h</math> to the identity element of <math>K</math>. Hence, these two maps are unequal.</toggledisplay>
			\|-
			\| 3 \|\| There are only finitely many homomorphisms from <math>G</math> to <math>K</math> \|\| \|\| \|\| <toggledisplay>A homomorphism of groups is completely specified by where the generators go. Since <math>G</math> has a finite generating set and <math>K</math> is finite, there are only finitely many possibilities for a homomorphism, bounded by <math>\|K\|^s</math> where <math>s</math> is the minimum size of generating set for <math>G</math>.</toggledisplay>
			\|-
			\| 4 \|\| We have the required contradiction \|\| \|\| Steps (2), (3) \|\| <toggledisplay>Step (2) gives infinitely many pairwise distinct homomorphisms from <math>G</math> to <math>K</math>, Step (3) asserts that there are only finitely many.
			\|}

	==External links==		==External links==

	* [http://mathoverflow.net/questions/22801/ MathOverflow discussion]		* [http://mathoverflow.net/questions/22801/ MathOverflow discussion]

Vipul: /* Proof */

2010-05-23T02:30:32Z

Proof

← Older revision		Revision as of 02:30, 23 May 2010
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	\| [[Hopfian group]] \|\| every [[surjective endomorphism]] of the group is an [[automorphism]]		\| [[Hopfian group]] \|\| every [[surjective endomorphism]] of the group is an [[automorphism]]
	\|}		\|}

			==Facts used==

			# [[uses::Finitely generated implies finitely many homomorphisms to any finite group]]
			# [[uses::Residually finite and finitely many homomorphisms to any finite group implies Hopfian]]

	==Proof==		==Proof==

			===Proof from given facts===

			The proof follows directly by piecing together facts (1) and (2).

			===Hands-on proof===

	'''Given''': A finitely generated group <math>G</math> that is also residually finite. A surjective homomorphism <math>\varphi:G \to G</math>.		'''Given''': A finitely generated group <math>G</math> that is also residually finite. A surjective homomorphism <math>\varphi:G \to G</math>.
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	# There is a surjective homomorphism <math>\alpha:G \to K</math> for some finite group <math>K</math> such that <math>\alpha(g)</math> is not the identity element: <toggledisplay>Since <math>G</math> is residually finite, and <math>g</math> is a non-identity element, there is a normal subgroup <math>N</math> of finite index in <math>G</math> such that <math>g \notin N</math>. Let <math>K = G/N</math> and <math>\alpha:G \to K</math> be the quotient map.</toggledisplay>		# There is a surjective homomorphism <math>\alpha:G \to K</math> for some finite group <math>K</math> such that <math>\alpha(g)</math> is not the identity element: <toggledisplay>Since <math>G</math> is residually finite, and <math>g</math> is a non-identity element, there is a normal subgroup <math>N</math> of finite index in <math>G</math> such that <math>g \notin N</math>. Let <math>K = G/N</math> and <math>\alpha:G \to K</math> be the quotient map.</toggledisplay>
	# All the homomorphisms <math>\alpha \circ \varphi^n</math> are distinct homomorphisms from <math>G</math> to <math>K</math>: <toggledisplay>Suppose <math>m < n</math>. We need to show that <math>\alpha \circ \varphi^m \ne \alpha \circ \varphi^n</math>. Since <math>\varphi</math> is surjective, <math>\varphi^m</math> is also surjective. In particular, there is <math>h \in G</math> such that <math>\varphi^m(h) = g</math>. Since <math>n > m</math> and <math>g</math> is in the kernel of <math>\varphi</math>, <math>\varphi^n(g)</math> is the identity element. Thus, <math>\alpha \circ \varphi^m</math> sends <math>h</math> to a non-identity element of <math>K</math> but <math>\alpha \circ \varphi^n</math> sends <math>h</math> to the identity element of <math>K</math>. Hence, these two maps are unequal.</toggledisplay>		# All the homomorphisms <math>\alpha \circ \varphi^n</math> are distinct homomorphisms from <math>G</math> to <math>K</math>: <toggledisplay>Suppose <math>m < n</math>. We need to show that <math>\alpha \circ \varphi^m \ne \alpha \circ \varphi^n</math>. Since <math>\varphi</math> is surjective, <math>\varphi^m</math> is also surjective. In particular, there is <math>h \in G</math> such that <math>\varphi^m(h) = g</math>. Since <math>n > m</math> and <math>g</math> is in the kernel of <math>\varphi</math>, <math>\varphi^n(g)</math> is the identity element. Thus, <math>\alpha \circ \varphi^m</math> sends <math>h</math> to a non-identity element of <math>K</math> but <math>\alpha \circ \varphi^n</math> sends <math>h</math> to the identity element of <math>K</math>. Hence, these two maps are unequal.</toggledisplay>
	# There are only finitely many homomorphisms from <math>G</math> to <math>K</math>: <toggledisplay>A homomorphism of groups is completely specified by where the generators go. Since <math>G</math> has a finite generating set and <math>K</math> is finite, there are only finitely many possibilities for a homomorphism, bounded by <math>\|K\|^s</math> where <math>s</math> is the minimum size of generating set for <math>G</math>.</toggledisplay>		# There are only finitely many homomorphisms from <math>G</math> to <math>K</math> : This follows from fact (1). <toggledisplay>A homomorphism of groups is completely specified by where the generators go. Since <math>G</math> has a finite generating set and <math>K</math> is finite, there are only finitely many possibilities for a homomorphism, bounded by <math>\|K\|^s</math> where <math>s</math> is the minimum size of generating set for <math>G</math>.</toggledisplay>
	# We have the required contradiction from (2) and (3).		# We have the required contradiction from (2) and (3).

Vipul: /* Definitions used */

2010-05-13T15:18:08Z

Definitions used

← Older revision		Revision as of 15:18, 13 May 2010
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	\| [[Finitely generated group]] \|\| has a finite [[generating set of a group\|generating set]]		\| [[Finitely generated group]] \|\| has a finite [[generating set of a group\|generating set]]
	\|-		\|-
	\| [[Residually finite group]] \|\| every non-identity element is outside some [[normal subgroup of finite index]]		\| [[Residually finite group]] \|\| every non-identity element is outside some [[normal subgroup of finite index]]. In particular, there is a surjective homomorphism to a finite group such that the given non-identity element is ''not'' in the kernel of the homomorphism.
	\|-		\|-
	\| [[Hopfian group]] \|\| every [[surjective endomorphism]] of the group is an [[automorphism]]		\| [[Hopfian group]] \|\| every [[surjective endomorphism]] of the group is an [[automorphism]]

Vipul: /* Proof */

2010-05-13T15:16:20Z

Proof

← Older revision		Revision as of 15:16, 13 May 2010
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	==Proof==		==Proof==

	~~{{fillin}}~~		'''Given''': A finitely generated group <math>G</math> that is also residually finite. A surjective homomorphism <math>\varphi:G \to G</math>.

			'''To prove''': <math>\varphi</math> is an automorphism of <math>G</math>.

			'''Proof''': We prove this by contradiction. Suppose <math>\varphi</math> is not an automorphism of <math>G</math>. Then, since it is surjective, there exists an non-identity element <math>g</math> in its kernel.

			# There is a surjective homomorphism <math>\alpha:G \to K</math> for some finite group <math>K</math> such that <math>\alpha(g)</math> is not the identity element: <toggledisplay>Since <math>G</math> is residually finite, and <math>g</math> is a non-identity element, there is a normal subgroup <math>N</math> of finite index in <math>G</math> such that <math>g \notin N</math>. Let <math>K = G/N</math> and <math>\alpha:G \to K</math> be the quotient map.</toggledisplay>
			# All the homomorphisms <math>\alpha \circ \varphi^n</math> are distinct homomorphisms from <math>G</math> to <math>K</math>: <toggledisplay>Suppose <math>m < n</math>. We need to show that <math>\alpha \circ \varphi^m \ne \alpha \circ \varphi^n</math>. Since <math>\varphi</math> is surjective, <math>\varphi^m</math> is also surjective. In particular, there is <math>h \in G</math> such that <math>\varphi^m(h) = g</math>. Since <math>n > m</math> and <math>g</math> is in the kernel of <math>\varphi</math>, <math>\varphi^n(g)</math> is the identity element. Thus, <math>\alpha \circ \varphi^m</math> sends <math>h</math> to a non-identity element of <math>K</math> but <math>\alpha \circ \varphi^n</math> sends <math>h</math> to the identity element of <math>K</math>. Hence, these two maps are unequal.</toggledisplay>
			# There are only finitely many homomorphisms from <math>G</math> to <math>K</math>: <toggledisplay>A homomorphism of groups is completely specified by where the generators go. Since <math>G</math> has a finite generating set and <math>K</math> is finite, there are only finitely many possibilities for a homomorphism, bounded by <math>\|K\|^s</math> where <math>s</math> is the minimum size of generating set for <math>G</math>.</toggledisplay>
			# We have the required contradiction from (2) and (3).

	==External links==		==External links==

	* [http://mathoverflow.net/questions/22801/ MathOverflow discussion]		* [http://mathoverflow.net/questions/22801/ MathOverflow discussion]

← Older revision		Revision as of 00:55, 23 January 2012
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	{{group property implication\|		{{group property implication\|
	stronger = finitely generated residually finite group\|		stronger = finitely generated residually finite group\|
	weaker = Hopfian group}}		stronger relevance = 1\|
			weaker = Hopfian group\|
			weaker relevance = 1}}

	[[difficulty level::3\| ]]		[[difficulty level::3\| ]]