Example of a subgroup What's an example of a subgroup H with $|H| = 12$ in $D_5 × Z_{30}$. Does it have to be normal or cyclic?
Attempt: If $H = D_m × Z_n$ then $2mn = 12$ with $m = 1,...,12$ and $n = 1,...,30$ so the only option is $m = 0$; $n = 12$ which is impossible. $|D_5 × Z_{30}| = 10(30) = 300 = 25(12)$. If $Z(D_5 × Z_{30}) ≥ |H|$ then $H$ must be cyclic. Is this true? How do I check if it's normal?
 A: As a partial answer. I am not sure it has to be cyclic. An example would be, the group generated by $\{(I, 5), (R, 0)\}$ where $I$ and $R$ are resp. the identity and a reflection in $D_5$. Since $(R, 0)$ is of order $2$ and $(I, 5)$ of order 6. But it's probable I am mistaken, since I don't know much about group theory.
And I think you can prove this group isn't normal either in taking $\theta$ as the rotation of one fifth and prove that $\theta R \theta^{-1}$ is neither $R$ or $I$ so $(\theta, 0)(R, 0)(\theta, 0)^{-1}$ isn't in $H$.
A: As $D_5$ has no elements of order 3, 4, 6 or 12, it should be clear that $H$ will need to be generated by an element of order 2 in $D_5$ and an element of order $6$ in $Z_{30}$. Write $G=D_5\times Z_{30}$ as  $$G=\{(a,b): a\in D_5; b\in Z_{30}\}$$with $D_5$ written multiplicatively and $Z_{30}$ written additively, and having identity $(e,0)$ and with the obvious componentwise group multiplication. Let $d$ be a fixed element of order 2 in $D_5$, and put $$H=\{(a,b):a\in \langle d\rangle; b\in \langle 5 \rangle \}$$ Check that $|H|=12$ and try to determine its structure. You might want to write out some products explicitly to see what is going on. 
To determine if $H$ is normal, try to find representative elements of $G$ of various "types" and let them act by conjugation on elements of $H$. It is not necessary to write out all of $G$ of course, but you should spend some time exploring how $G$ "works". 
