Please tell me how to formalize my thought if my thought is not wrong. ("obvious by symmetry", "symmetric group")

Question

Let $S_4$ be the symmetric group of order $4$.
$S_4=\{\operatorname{id},(1 2), (1 3), (1 4), (2 3), (2 4), (3 4), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3), (1 2 3), (1 3 2), (1 2 4), (1 4 2), (1 3 4), (1 4 3), (2 3 4), (2 4 3), (1 2 3 4), (1 2 4 3), (1 3 2 4), (1 3 4 2), (1 4 2 3), (1 4 3 2)\}.$

Let $H:=\{(1 2)(3 4), (1 3)(2 4), (1 4)(2 3)\}$.
Prove that $\phi(a)\in H$ for any $a\in H$ and for any $\phi\in\operatorname{Aut}(S_4)$

If I must solve the above problem, do I need to calculate the values of $\phi((1 2)(3 4)),\phi((1 3)(2 4)),\phi((1 4)(2 3))$ for all $\phi\in\operatorname{Aut}(S_4)$?

By symmetry, I think each element of $H$ has the same algebraic properties in $S_4$.
And $\phi\in\operatorname{Aut}(S_4)$ is an automorphism.
So, I think it is obvious that $\phi(a)\in H$ holds for any $a\in H$ and for any $\phi\in\operatorname{Aut}(S_4)$.

But I don't know how to formalize my thought.

Please tell me how to formalize my thought if my thought is not wrong. (I used $S_4$ as an example.)

Answer 1

You can argue this way.

Note that $\phi$ fixes $\{e\}$. So we can replace your $H$ by the subgroup $H=\{e, (12)(34), (13)(24), (14)(23)\}$: if $\phi$ fixes this subgroup it fixes your $H$. So from now on $H$ is the subgroup.

The subgroup $H$ is normal, since conjugation in $S_n$ preserves cycle-type. Now put $K=\phi(H)$, and check that $K$ is also normal. (Automorphisms preserve normality of subgroups).

As $H$ and $K$ are normal so is $H\cap K$.

We can't have $|H\cap K|=1$ as $16$ doesn't divide $24$.

We can't have $|H\cap K|=2$. Without loss take the case when $H\cap K=\{e,(12)(34)\}$. Then $(123)^{-1}(H\cap K)(123)\ne H\cap K$.

So $|H\cap K|=4$ and we are done, $H=K$.

Answer 2

Automorphisms preserve elements' order and automorphism group naturally acts on the set of the conjugacy classes. So, every automorphism must send the conjugacy class of the double transpositions (such is $H$) to itself, because the conjugacy class of the transpositions (the only other one whose elements have order $2$) has six elements, while $H$ has three. By a stronger condition (distinct conjugacy classes have distinct orders of their elements), the claim holds for every other conjugacy class of $S_4$ (and for all those of $S_3$). The case $S_5$ can be dealt with likewise. The first case where this argument is not conclusive is $S_6$: here, for example, the two conjugacy classes of the transpositions and of the triple transpositions have the same size ($15$). In principle, some (non-inner) automorphism might swap these two classes. And actually there are which do.