Prove that $\Bbb Z/6\Bbb Z\cong\Bbb Z/3\Bbb Z \times\Bbb Z/2\Bbb Z$

Define the map $f : \Bbb Z/6\Bbb Z\to\Bbb Z/3\Bbb Z \times\Bbb Z/2\Bbb Z$ by $f([a]_6) = ([a]_3, [a]_2)$ where $[a]_n$ denotes the residue class of $a \in\Bbb Z$ modulo $n$. Prove that $f$ is an isomorohism. Try to prove that if $\gcd(m, n) = 1$, then the groups $\Bbb Z/mn\Bbb Z$ and $\Bbb Z/m\Bbb Z\times\Bbb Z/n\Bbb Z$ are isomorphic.

Can anyone help with this question?

I have prove only that $\gcd(m,n)=1$.

• Are you aware of the first isomorphism theorem? How do you normally show a map is an isomorphism? – Dave Feb 23 '18 at 16:07
• before to do the isomorphic theorem I think I have to prove first that this equation is homomorphic. – soc Feb 23 '18 at 16:09
• for example f(a+b)=f(a)+f(b) – soc Feb 23 '18 at 16:09
• I suppose I should ask this as well: are you trying to show they are isomorphic as rings or groups? – Dave Feb 23 '18 at 16:11
• I am trying to show that they are isomorphic as group – soc Feb 23 '18 at 16:12

$f:\mathbb{Z_6}\to\mathbb{Z_3}\times\mathbb{Z_2}$ with $f([a]_6)=([a]_3,[a]_2)$

• $f$ is a homomorphism of groups:

$f([a]_6+[b]_6)=f([a+b]_6)=([a+b]_3,[a+b]_2)=([a]_3+[b]_3,[a]_2+[b]_2)=([a]_3,[a]_2)+([b]_3,[b]_2)=f([a]_6)+f([b]_6)$

• $f$ is $1-1$:

$f([a]_6)=f([b]_6)\Rightarrow ([a]_3,[a]_2)=([b]_3,[b]_2)\Rightarrow [a]_3=[b]_3$ and $[a]_2=[b]_2\Rightarrow 3|a-b$ and $2|a-b$ and $gcd(3,2)=1$ so $6|a-b\Rightarrow[a]_6=[b]_6$

• $f$ is onto:

$f$ is $1-1$ and $|\mathbb{Z_6}|=|\mathbb{Z_3}\times\mathbb{Z_2}|=6\Rightarrow f$ is also onto

The fact that $\mathbb Z/N \to\mathbb Z/n \times \mathbb Z/m$ where $N=m \cdot n$, and both coprime is known as the chinese remainder theorem.

The fact that it is injective, is an application of the first isomorphism theorem for groups, since the map $\varphi:\mathbb Z \to \mathbb Z/m \times \mathbb Z/n$ given by $a \mapsto(a,a)$ is in the kernel if and only if $m \mid a$ and $n \mid a$ if and only if $N \mid a$, or in other words, $a \in N \mathbb Z$, so $\mathbb Z/\ker \varphi=\mathbb Z/N$.

In your case, you can get your hands dirty:

hint for surjectivity: for surjectivity, show that there exists some $x \in \mathbb Z/6$ so that $x \equiv a_1 \mod 2$ and $x \equiv a_2 \mod 3$ for any choice of $(a_1,a_2) \in \mathbb Z/2 \times \mathbb Z/3$. One way to approach this, is to find integers $m,n$ so that $2m+3n=1$ and set $x=3a_1n+2a_2m$ and reduce mod $6$.

Hint for injectivity: reproduce the argument given in the second paragraph, but specialize it for your problem.

Hint if you're feeling lazy: take whichever problem (injectivity/ surjectivity) you find easier, and note that the two groups have the same cardinality, so either condition implies the other.