continuation of problem subgroup isomorphic

Question

This subject is associated with Does group $A_6$ contain subgroup isomorphic with $S_4$

Hint: Take $S_4$ on the first four elements of the set permuted by $A_6$. Some of these are odd permutations. Can you see an easy way to convert them into even permutations which are contained in $A_6$? Can you do this in a consistent way which doesn't disturb the $S_4$ structure?

I do my best to understand it. Isomorphism is a function betwenn groups. In this case we would like to find subgroup of $A_6$ isomorphic to $S_4$.

$A_6$ work with $\{1,2,3,4,5,6\}$ and all of permutations are even. You give me a hint. I get any of permutation in $S_4$, for example $(13)(24)$. Thus, when I work on firstly four elements: We get: $(13)(24)(5)(6)$

Is it what you mean ?

Answer 1

We claim that: $$ H = A_4 \cup \{(5,6)\alpha \mid \alpha \in S_4 \setminus A_4 \} $$ is our desired subgroup. To see this, first we need to show that $H \leq A_6$:

• $\boxed{\emptyset \neq H \subseteq A_6}$: Can you see why each element of $H$ is an even permutation?
• $\boxed{\alpha,\beta \in H \implies \alpha^{-1}\beta \in H}$: Can you see why the composition of any two permutations in $H$ will either be in $A_4$ or have the form $(5,6)\alpha$ for some odd $\alpha$?

We also need to show that the function $f\colon S_4 \to H$ defined by: $$ f(\alpha) = \begin{cases} \alpha &\text{if } \alpha \in A_4 \\ (5,6)\alpha &\text{if } \alpha \in S_4 \setminus A_4 \\ \end{cases} $$ is an isomorphism:

• $\boxed{f \text{ preserves structure}}$: Can you see why $f(\alpha)f(\beta) = f(\alpha\beta)$?
• $\boxed{f \text{ is injective}}$: Can you see why $f(\alpha) = f(\beta) \implies \alpha = \beta$?
• $\boxed{f \text{ is surjective}}$: This follows immediately by construction.