Isomorphic iff potentially conjugate in finite: Difference between revisions

Revision as of 21:41, 6 March 2009

Statement

With one isomorphism

Suppose $G$ is a finite group and $σ : H \to K$ is an isomorphism between two subgroups of $G$ . Then, there exists a finite group $S$ containing $G$ and an element $g \in S$ such that $g x g^{- 1} = σ (x)$ for any $x \in H$ .

With multiple isomorphisms

Suppose $G$ is a finite group and $σ_{i} : H_{i} \to K_{i}$ is a collection of isomorphisms between subgroups $H_{i}, K_{i}$ of $G$ , with $i \in I$ . Then, there exists a finite group $S$ containing $G$ and elements $g_{i} \in S$ such that $g_{i} x g_{i}^{- 1} = σ (x)$ for any $x \in H_{i}$ .

Related facts

Isomorphic iff potentially conjugate: The general version for infinite groups. Note that the proof outlined in the finite case does not work directly for infinite groups, because the finite groups version uses the fact that isomorphic subgroups have the same index.

Proof

With one isomorphism

Let $S$ be the symmetric group on the set $G$ with $G$ viewed as a subgroup of $S$ via the left multiplication action (this is the embedding given by Cayley's theorem). Under this embedding, the isomorphism $σ : H \to K$ lifts to an inner automorphism given by conjugation by a permutation $α$ constructed as follows.

The restriction of $α$ to $H$ is $σ$ . Further, pick coset representatives for $H, K$ in $G$ such that the identity element is the coset representative of the subgroups themselves. Now, define $α$ as a bijection between these sets of coset representatives such that $α$ sends the identity element to the identity element. Finally, for any element of the form $x h$ , with $x$ a coset representative of $H$ and $h \in H$ , define $α (x h) = α (x) σ (h)$ .

We verify that this works. PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

With multiple isomorphisms

Since $G$ is finite, so is the set $I$ of isomorphisms. Without loss of generality, $I = {1, 2, 3, \dots, n}$ . The rough idea is that each time we take the symmetric group on the set, we can convert one isomorphism to an inner automorphism. Thus, repeating the process $n$ times yields a finite group in which all the isomorphisms are realized as inner automorphisms.

The new finite group has order $f^{n} (| G |)$ , where $f$ is the factorial function. This grows very rapidly with $n$ for $| G | > 2$ . For instance, $f^{2} (6)$ is $720!$ which is a number with several thousand digits.

@@ Line 3: / Line 3: @@
 ===With one isomorphism===
-Suppose <math>G</math> is a [[finite group]] and <math>\sigma:H \to K</math> is an isomorphism between two subgroups of <math>G</math>. Then, there exists a finite group <math>S</math> containing <math>G</math> and an element <math>g \in S</math> such that <math>gxg^{-1} = \sigma(g)</math> for any <math>x \in H</math>.
+Suppose <math>G</math> is a [[finite group]] and <math>\sigma:H \to K</math> is an isomorphism between two subgroups of <math>G</math>. Then, there exists a finite group <math>S</math> containing <math>G</math> and an element <math>g \in S</math> such that <math>gxg^{-1} = \sigma(x)</math> for any <math>x \in H</math>.
 ===With multiple isomorphisms===
-{{fillin}}
+Suppose <math>G</math> is a [[finite group]] and <math>\sigma_i:H_i \to K_i</math> is a collection of isomorphisms between subgroups <math>H_i, K_i</math> of <math>G</math>, with <math>i \in I</math>. Then, there exists a finite group <math>S</math> containing <math>G</math> and elements <math>g_i \in S</math> such that <math>g_ixg_i^{-1} = \sigma(x)</math> for any <math>x \in H_i</math>.
 ==Related facts==
-* [[Isomorphic iff potentially conjugate]]: The general version for infinite groups.
+* [[Isomorphic iff potentially conjugate]]: The general version for infinite groups. Note that the proof outlined in the finite case does not work directly for infinite groups, because the finite groups version uses the fact that isomorphic subgroups have the same index.
 ==Proof==
+===With one isomorphism===
 Let <math>S</math> be the symmetric group on the set <math>G</math> with <math>G</math> viewed as a subgroup of <math>S</math> via the left multiplication action (this is the embedding given by [[Cayley's theorem]]). Under this embedding, the isomorphism <math>\sigma:H \to K</math> lifts to an inner automorphism given by conjugation by a permutation <math>\alpha</math> constructed as follows.
@@ Line 20: / Line 23: @@
 We verify that this works. {{fillin}}
+===With multiple isomorphisms===
+Since <math>G</math> is finite, so is the set <math>I</math> of isomorphisms. Without loss of generality, <math>I = \{ 1,2,3,\dots,n\}</math>. The rough idea is that each time we take the symmetric group on the set, we can convert one isomorphism to an inner automorphism. Thus, repeating the process <math>n</math> times yields a finite group in which all the isomorphisms are realized as inner automorphisms.
+The new finite group has order <math>f^n(|G|)</math>, where <math>f</math> is the factorial function. This grows very rapidly with <math>n</math> for <math>|G| > 2</math>. For instance, <math>f^2(6)</math> is <math>720!</math> which is a number with several thousand digits.