# Showing an isomorphism exists between either $\mathbb Z_4$ or $\mathbb Z_2 \times \mathbb Z_2$

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.

• 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 $\mathbb{Z}_2 \times \mathbb{Z}_2$. – user296602 Oct 21 '18 at 22:03
• So then $H$ is isomorphic to $\mathbb Z_2 \times \mathbb Z_2$ then, correct? @T.Bongers – Propaloo Oct 21 '18 at 22:13
• Just a note: your group $H$ needs to include the identity element! – 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.