If there is a bijection $f: X\rightarrow Y$, prove that there exists an isomorphism $\phi :S_X\rightarrow S_Y$

If there is a bijection $f: X\rightarrow Y$, prove that there exists an isomorphism $\phi :S_X\rightarrow S_Y$.

Here $S_X$ denotes the group of all permutations of $X$, i.e., the bijections $X\to X$ and the group operation is composition.

I know that $X$ and $Y$ have the same cardinality because of the bijection. I define $\phi:S_X \rightarrow S_Y$ by $\phi : a \rightarrow f\circ a\circ f^{-1}$ Then I suppose $\phi^{-1}$ can be defined as $b\rightarrow f^{-1} \circ b\circ f$.

I think this is close but I haven't showed that $\phi$ is a homomorphism. But once I do I can show isomorphism.

• By $S_X$ you mean the group of permutations over $X$?all bijections from $X$onto itself? – BigM Feb 20 '15 at 5:01
• @BigM Yes that's what I meant. – grayQuant Feb 20 '15 at 5:46

$\phi(a\circ b)= f\circ (a\circ b)\circ f^{-1}=(f\circ a\circ f^{-1})\circ (f \circ b\circ f^{-1})=\phi(a)\circ \phi(b)$

• Why $f\circ (a\circ b)\circ f^{-1}=(f\circ a\circ f^{-1})\circ (f \circ b\circ f^{-1})$? – Alan Wang Aug 12 '15 at 3:59
• Because $f$ is a bijection. – BigM Aug 13 '15 at 5:13

1. What is a permutation? It is bijection from a set to itself.

2. The composition of two bijection functions is bijection.

• Your definition of permutation is not correct. It needs to be bijective. – Tobias Kildetoft Feb 20 '15 at 8:24
• @TobiasKildetoft Usually $S_X$ denotes the symmetric group on a finite set X. The injection from a finite set to itself is bijection. :) – user80225 Feb 20 '15 at 8:36
• I disagree with the "usually". – Tobias Kildetoft Feb 20 '15 at 8:37
• @TobiasKildetoft +1 for disagree. Anyway, I will change it. Thanks – user80225 Feb 20 '15 at 8:38
• @user31009 In a group, the elements have inverses. So how do you invert injective but non-surjective maps on an infinite set? – Christoph Feb 20 '15 at 8:46