# In $D_4$, find $H_1$ isomorphic to $\mathbb{Z}_4$ and $H_2$ isomorphic to the Klein Four Group with $D_4/H_1$ isomorphic to $D_4/H_2$.

a) In $D_4$, find $H_1$ isomorphic to $\mathbb{Z}_4$ and $H_2$ isomorphic to $V$, the Klein Four group, with $D_4/H_1$ isomorphic to $D_4/H_2$.

b) In $D_4$, find subgroups $H$ and $K$ with $H$ normal in $K$ and $K$ normal in $D_4$ but $H$ not not normal in $D_4$.

-and in this I'm just confused.

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What do you mean by 4Z/Z, and how is it inside D4? Perhaps you mean Z/4Z? But you're required to find a subgroup of D4 that's iso to Z4, not a quotient of the integers. I think it would benefit you to just list out every possible subgroup of D4 and then check their isomorphism types, then check which subgroups are normal in which other subgroups (including the entire group itself). –  anon Dec 14 '12 at 16:59
I changed $4\mathbb{Z}/\mathbb{Z}$ to $\mathbb{Z}/4\mathbb{Z}$ as this was almost certainly a typo. I don't know what you mean by $8\mathbb{Z}/\mathbb{Z}_2$ though. –  Matthew Pressland Dec 14 '12 at 16:59
@MattPressland Presumably $2\Bbb Z/8\Bbb Z$... –  anon Dec 14 '12 at 17:00
@anon That was my best guess, but it seems like an odd thing to say, so I didn't want to change it to that without the OP's approval. –  Matthew Pressland Dec 14 '12 at 17:00
(And 4Z/Z is just a typo rather than an odd thing to say? :P) –  anon Dec 14 '12 at 17:01

Some stuff that will help here: $D_4$ has a cyclic subgroup of order $4$ and all other elements have order $2$ (the subgroup of order $4$ is the one given by the rotations).
So we already have a subgroup of order $4$ which is cyclic.
To find one of order $4$ which is not cyclic, we note that the element of order $2$ in the above mentioned cyclic subgroup is central, so if we let $x$ be this element and $y$ be any other element of order $2$, then $\{e,x,y,xy\}$ is a subgroup of order $4$ which is not cyclic.
The part about the quotients by these two subgroups being isomorphic is automatic, as they both have order $2$, and there is only one group of order $2$.
For the second part, we can take the non-cyclic subgroup from above and the subgroup of that generated by $y$. Since $y$ has order $2$ and is not central, the subgroup generated by $y$ cannot be normal.