Finding normal subgroups of G given a homomorphism from $G$ onto $Z_6 \oplus Z_2$ with $|Ker\phi|=5$ This problem is from Gallian's Contemporary Abstract Algebra, 9TH edition, p.207
Suppose that $\phi$ is a homomorphism from a group $G$ onto $Z_6 \oplus Z_2$ and that the kernel of $\phi$ has order 5. Explain why G must have normal subgroups of orders 5, 10, 15, 20, 30 and 60.
I know that $\phi$ is a 5-to-1 mapping from G to $Z_6 \oplus Z_2$. But I don't know how to use the following theorem to find the order of the normal subgroups:
If $K'$ is a normal subgroup of $G'$, then $\phi^{-1}(K')=\{k \in G : \phi(k) \in K'\}$ is a normal subgroup of $G$.
 A: You have a group $\;G\;$ such that there exists an epimorphism $\;\phi: G\to K:=\Bbb Z_6\oplus\Bbb Z_2\;$ with $\;H:=\ker\phi\;$ of order $\;5\;$ , and this means by the first isomorphism theorem that
$$G/H\cong K\implies |G|=|H|\cdot|K|=60$$
and we already have normal subgroups of $\;G\;$ of order  $\;5,\,60\;$ .
But $\;K\;$ is abelian and thus all its subgroups are normal, which means (by the correspondence theorem) that their inverse images under $\;\phi\;$ are normal subgroups of $\;G\;$ , so for example: there exists a normal subgroup $\;\overline M\le K\;$ of order $\;2\;$, which we can write as to  $\;\overline M=M/H\;$ , for some subgroup $\;H\le M\le G\;$, and the corr. theorem also tells us that
$$6=\frac{12}2=[K:\overline M]=\left|\left(G/H\right)/\left(M/H\right)\right|=|G/M|\implies |M|=10$$
and there you have your normal subgroup of order $\;10\;$ . Take it from here now.
A: Use what you know about $\mathbb Z_6\oplus\mathbb Z_2$...  In particular,  there are normal subgroups of indices $1,2,3,4,6$ and $12$.  The preimages of these subgroups are the normal subgroups you're looking for...  (by the theorem you mentioned).
