Is $(\Bbb{Z}/12\Bbb{Z})/(3 \Bbb{Z}/12\Bbb{Z}) \times 3 \Bbb{Z}/12\Bbb{Z} \cong \Bbb{Z}/12\Bbb{Z}$? Let $G$ be the group $$G=\Bbb{Z}/12\Bbb{Z} = \{\overline{x}:\overline{x} = \overline{0},\overline{1},\overline{2}...,\overline{11}\}$$ and $N$ be the subgroup  $$N = 3 \Bbb{Z}/12\Bbb{Z} = \{\overline{3}\overline{x}:\overline{x} \in \Bbb{Z}/12\Bbb{Z}\} = \{\overline{0},\overline{3},\overline{6},\overline{9}\}$$
My qestion is, is the product of the quotient group $G/N$ with the normal group $N$, isomorphic to $G$? I.e., is $$G/N \times N \cong G\text{?}$$
I think that $G/N = \{\{\overline{0},\overline{3},\overline{6},\overline{9}\},\{\overline{1},\overline{4},\overline{7},\overline{10}\},\{\overline{2},\overline{5},\overline{8},\overline{11}\}\} = \{\overline{0}_3,\overline{1}_3,\overline{2_3}\}$, where $\overline{x}_3 = \{x+3n :n \in \Bbb{Z}\}$
I think the answer to the isomorphism question is yes, because I think you can construct the isomorphism $f : G/N \times N \to G$ given by $f(\overline{x}_3,\overline{y}) = \overline{x}+\overline{y}$, and i think this is an isomorphism. I'm not sure how to prove it though. I'm very new to quotient groups and I don't think I understand them, so if anyone could explain to me if there's an easier way of showing to groups are isomorphic, or if I've gotten the quotient group wrong I sould appreciate it. Thanks!
 A: Hint:
The third isomorphism theorem asserts that we have a canonical isomorphism:
$$(\mathbf Z/12\mathbf Z)\!\bigm/\!(3\mathbf Z/12\mathbf Z)\simeq\mathbf Z/3\mathbf Z.$$
On the other hand we have a well-defined isomorphism:
\begin{align}
3\mathbf Z/12\mathbf Z &\longrightarrow\mathbf Z/4\mathbf Z \\
3k+12\mathbf Z=3(k+4\mathbf Z)&\longmapsto k+4\mathbf Z
\end{align}
Finally we have the isomorphism from the Chinese remainder theorem:
$$\mathbf Z/12\mathbf Z\simeq \mathbf Z/4\mathbf Z \times  \mathbf Z/3\mathbf Z. $$
Can you proceed from here?
A: It is not true in general that $G/N\times N\cong G$, but it is true in this case.  I am not going to give the most efficient way to prove the isomorphism, but instead I will give a way which might give more of an idea for how to work with quotient groups computationally.
The elements of $\mathbb{Z}/12\mathbb{Z}$ are cosets of the subgroup $12\mathbb{Z}$.  Since the group is additive, we can write an element of the quotient as $a+12\mathbb{Z}$ for $a\in\mathbb{Z}$, where this represents the set $\{a+12k:k\in\mathbb{Z}\}$.  The addition rule is $(a+12\mathbb{Z})+(b+12\mathbb{Z})=(a+b)+\mathbb{Z}$, which follows from agreeing that the sum of two sets is the set of all sums.
Let $f:\mathbb{Z}/12\mathbb{Z}\times3\mathbb{Z}/12\mathbb{Z}\to\mathbb{Z}/12\mathbb{Z}$ be defined by $f(n+12\mathbb{Z},3m+12\mathbb{Z})=(4n+\mathbb{Z})+(3m+12\mathbb{Z})=4n+3m+12\mathbb{Z}$.  (It turns out just summing the cosets is not quite what you want for this particular argument; you will end up with a more complicated quotient than you would like.*)
The kernel of this map is determined by the set of all $(n,m)$ such that $4n+3m\in12\mathbb{Z}$, since $12\mathbb{Z}$ is the zero element in $\mathbb{Z}/12\mathbb{Z}$.  Using the Chinese remainder theorem, being $0$ mod $12$ is the same as being $0$ mod both $3$ and $4$.  Hence, $n=3a$ and $m=4b$ for some $a,b\in\mathbb{Z}$, so the kernel is parameterized by $(n,m)=(3a,4b)$, which corresponds in the cartesian product to pairs $(3a+12\mathbb{Z},3(4b)+12\mathbb{Z})=(3a+12\mathbb{Z},12\mathbb{Z})$.  Hence, the kernel is exactly $3\mathbb{Z}/12\mathbb{Z}\times\{0\}$.  By the first isomorphism theorem, $\overline{f}:(\mathbb{Z}/12\mathbb{Z})/(3\mathbb{Z}/12\mathbb{Z})\times 3\mathbb{Z}/12\mathbb{Z}\to\mathbb{Z}/12\mathbb{Z}$ is an isomorphism.
*In fact, the map you give is not well-defined.  Have you checked it was a homomorphism?
