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.
Best Answer
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 $f$.