Full invariance does not satisfy image condition: Difference between revisions

Latest revision as of 16:13, 22 May 2009

This article gives the statement, and possibly proof, of a subgroup property (i.e., fully invariant subgroup) not satisfying a subgroup metaproperty (i.e., image condition).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about fully invariant subgroup|Get more facts about image condition|

Statement

Suppose $G$ is a group, $K$ is a fully invariant subgroup of $G$ , and $φ : G \to H$ is a surjective homomorphism. Then, $φ (K)$ need not be fully invariant in $H$ .

Proof

Example of an Abelian group of prime-cube order

(This example uses additive notation).

Suppose $G$ is the direct product of a cyclic group $A$ of order $p$ and a cyclic group of order $B$ of order $p^{2}$ . Define:

$K = Ω_{1} (G)$ (see omega subgroups of a group of prime power order), i.e., $K$ is the subgroup comprising all the elements:

${x \in G ∣ p x = 0}$ .

$φ$ is the quotient map by the normal subgroup $℧^{1} (G)$ (see agemo subgroups of a group of prime power order), i.e., $φ$ is the quotient map by the subgroup:

$N : = {y \in G ∣ \exists x, p x = y}$ .

Observe that $K$ is fully invariant in $G$ (more generally, all omega subgroups are fully invariant). However, $φ (K)$ is a subgroup of order $p$ in $φ (G)$ which is elementary abelian of order $p^{2}$ -- hence $φ (K)$ is not fully invariant in $φ (G)$ .

Example of a non-abelian group of prime-cube order

Further information: Prime-cube order group:p2byp, Subgroup structure of prime-cube order group:p2byp

Let $p$ be an odd prime. Suppose $A$ is a cyclic group of order $p^{2}$ and $B$ is a cyclic group of order $p$ , with $B$ acting on $A$ via multiplication by $p + 1$ . Then, the semidirect product of $A$ by $B$ is a non-Abelian group of order $p^{3}$ . Call this group $P$ . Define $Ω_{1} (P)$ (see omega subgroups of a group of prime power order) as the subgroup generated by all elements of order $p$ in $P$ . By the fact that Omega-1 of odd-order class two p-group has prime exponent, $Ω_{1} (P)$ is a subgroup of prime exponent. This forces it to be a subgroup of order $p^{2}$ generated by the elements of $B$ and the multiples of $p$ in $A$ . All the omega subgroups are fully characteristic, so $Ω_{1} (P)$ is fully characteristic.

The center of $P$ , namely $Z (P)$ , simply comprises the multiples of $p$ in $A$ . Thus, in the quotient map $P \to P / Z (P)$ , the image of $Ω_{1} (P)$ is cyclic of order $p$ , while the whole group is elementary Abelian of order $p^{2}$ . Thus:

$Ω_{1} (P)$ is fully characteristic in $P$ .
The image of $Ω_{1} (P)$ in $P / Z (P)$ is not fully characteristic in $P / Z (P)$ .

@@ Line 1: / Line 1: @@
 {{subgroup metaproperty dissatisfaction|
-property = fully characteristic subgroup|
+property = fully invariant subgroup|
 metaproperty = image condition}}
 ==Statement==
-Suppose <math>G</math> is a group, <math>K</math> is a fully characteristic subgroup of <math>G</math>, and <math>\varphi:G \to H</math> is a surjective homomorphism. Then, <math>\varphi(K)</math> need not be fully characteristic in <math>H</math>.
+Suppose <math>G</math> is a group, <math>K</math> is a [[fully invariant subgroup]] of <math>G</math>, and <math>\varphi:G \to H</math> is a surjective homomorphism. Then, <math>\varphi(K)</math> need not be fully invariant in <math>H</math>.
 ==Proof==
-===Example of a non-Abelian group of prime-cubed order===
+===Example of an Abelian group of prime-cube order===
-{{further|[[Particular example::Prime-cube order group:p2byp]]}}
+(This example uses additive notation).
+Suppose <math>G</math> is the direct product of a cyclic group <math>A</math> of order <math>p</math> and a cyclic group of order <math>B</math> of order <math>p^2</math>. Define:
+* <math>K = \Omega_1(G)</math> (see [[omega subgroups of a group of prime power order]]), i.e., <math>K</math> is the subgroup comprising all the elements:
+<math>\{ x \in G \mid px = 0 \}</math>.
+* <math>\varphi</math> is the quotient map by the normal subgroup <math>\mho^1(G)</math> (see [[agemo subgroups of a group of prime power order]]), i.e., <math>\varphi</math> is the quotient map by the subgroup:
+<math>N := \{ y \in G \mid \exists x, px = y \}</math>.
+* Observe that <math>K</math> is fully invariant in <math>G</math> (more generally, all omega subgroups are fully invariant). However, <math>\varphi(K)</math> is a subgroup of order <math>p</math> in <math>\varphi(G)</math> which is elementary abelian of order <math>p^2</math> -- hence <math>\varphi(K)</math> is not fully invariant in <math>\varphi(G)</math>.
+===Example of a non-abelian group of prime-cube order===
+{{further|[[Particular example::Prime-cube order group:p2byp]], [[Subgroup structure of prime-cube order group:p2byp]]}}
+Let <math>p</math> be an odd prime.
 Suppose <math>A</math> is a cyclic group of order <math>p^2</math> and <math>B</math> is a cyclic group of order <math>p</math>, with <math>B</math> acting on <math>A</math> via multiplication by <math>p+1</math>. Then, the semidirect product of <math>A</math> by <math>B</math> is a non-Abelian group of order <math>p^3</math>. Call this group <math>P</math>. Define <math>\Omega_1(P)</math> (see [[omega subgroups of a group of prime power order]]) as the subgroup generated by all elements of order <math>p</math> in <math>P</math>. By the fact that [[Omega-1 of odd-order class two p-group has prime exponent]], <math>\Omega_1(P)</math> is a subgroup of prime exponent. This forces it to be a subgroup of order <math>p^2</math> generated by the elements of <math>B</math> and the multiples of <math>p</math> in <math>A</math>. All the omega subgroups are fully characteristic, so <math>\Omega_1(P)</math> is fully characteristic.