Doubts in $\Bbb Z_2 \times \Bbb Z_2$

Question

My question asks me to prove that whether $\Bbb Z_2 \times\Bbb Z_2$ a subgroup of $S_4$ or not.

I really cant get a clear picture but the only thing that I know is that if we can let $\Bbb Z_2 $ isomorphic to $H_1=\langle(a b)\rangle$ and $\Bbb Z_2$ isomorphic to $H_2=\langle(c d)\rangle$ .Then maybe I can show some isomorphism with some subgroup of $S_4$.[I am having a problem to explicitly write this]

There is also one thing that I know is that if I take $\Bbb Z_2$ as isomorphic to a the group $H_1=\langle(12)(34)\rangle$ and $H_2=\langle(23)(14)\rangle$ Then maybe some sort of isomorphism is possible. Can someone help out please by writing the proof or giving the general structure of the proof.

Answer 1

By Cayley's theorem, any group $G$ embeds into $S_G$ by left multiplication, say $\lambda$. If furtherly $G$ is finite, then $S_G\cong S_{|G|}$ via the isomorphism $\alpha\mapsto f^{-1}\alpha f$, where $f\colon\{1,\dots,|G|\}\to G$ is any bijection. So, yes, $G=\Bbb Z_2\times \Bbb Z_2$, like any other group of order $4$ (other, there is just the cyclic one, $\Bbb Z_4$), is isomorphic to a subgroup of $S_{|\Bbb Z_2\times \Bbb Z_2|}=S_4$. For instance, for:

\begin{alignat}{1} f(1)&=(0,0) \\ f(2)&=(0,1) \\ f(3)&=(1,0) \\ f(4)&=(1,1) \\ \end{alignat}

we get:

\begin{alignat}{1} f^{-1}\lambda(0,0)f&=() \\ f^{-1}\lambda(0,1)f&=(12)(34) \\ f^{-1}\lambda(1,0)f&=(13)(24) \\ f^{-1}\lambda(1,1)f&=(14)(23) \\ \end{alignat}

i.e. $\Bbb Z_2\times \Bbb Z_2\cong \{(),(12)(34),(13)(24),(14)(23)\}$. And similarly for other choices of $f$.

Answer 2

Yes, consider $\{e,(12)(34),(13)(24),(14)(23)\}$. It is isomorphic to the Klein four group, as all three nonidentity elements have order two.

Actually, there are four Klein four subgroups in $S_4$. We have $\{e,(12),(34),(12)(34)\},\{e,(13),(24),(13)(24)\}$ and $\{e,(14),(23),(14)(23)\}$. But only the first one above is normal.

As far as an isomorphism, you can map any of the nonidentity elements to any of the nonidentity elements of $V_4$, bijectively. So there are six.