For this question we are given the group $G = S_8$ and a subgroup $H$ $=$ {$(12)(58), (12)(36), (36)(58)$}. It wants us to find a group that is isomorphic to $H$. I know that $H$ has an order of $4$ so I was thinking that I could choose either the group $\mathbb Z_4$ or $\mathbb Z_2 \times \mathbb Z_2$ because they also have order $4$. Is this correct? Also, if I were to show one of these were isomorphic to $H$, what would my function $\phi$ be to show it is a bijection and operation preserving? Like for example, If I chose $\mathbb Z_4$, would it map $ \phi: H \to \mathbb Z_4$, or the other way around? Sorry, I am unsure about the approach to this problem. I appreciate any help given.

  • $\begingroup$ If $H$ has an element $h$ of order $4$, then $H$ is isomorphic to $\mathbb{Z}_4$ and you define $\phi$ by $\phi(h) = 1$ (with the obvious extension). Otherwise, $H$ is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$. $\endgroup$ – user296602 Oct 21 '18 at 22:03
  • $\begingroup$ So then $H$ is isomorphic to $\mathbb Z_2 \times \mathbb Z_2$ then, correct? @T.Bongers $\endgroup$ – Propaloo Oct 21 '18 at 22:13
  • $\begingroup$ Just a note: your group $H$ needs to include the identity element! $\endgroup$ – rogerl Oct 22 '18 at 1:08

Depends on what you mean by "choose". What's true is that there are (up to isomorphism) only two groups of order $4$, the two that you mentioned.

But, it's not that you can choose to which your $H$ is isomorphic, it's one or the other.

Notice that your $H$ has only elements of order $2$. Could it then be isomorphic to $\mathbb Z_4$ that has element of order $4$?

Well, no. So, choose any two elements of $H$ (apart from identity) and send one to $(1,0)$ and the other to $(0,1)$. This will be your isomorphism.


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