# Why is $\langle a,b\mid 2a=2b\rangle\cong\mathbb{Z}\oplus\mathbb{Z}_2$

May I ask, why is $$\langle a,b\mid 2a=2b\rangle\cong\mathbb{Z}\oplus\mathbb{Z}_2?$$

I am quite new with group generators. In this case, the coefficients are in $\mathbb{Z}$, and we are in the abelian case.

Thanks for any tips.

Update: I figured out an idea: Let $a=(1,1), b=(1,0)$ which are elements in $\mathbb{Z}\oplus\mathbb{Z}_2$. Then $a,b$ generate $\mathbb{Z}\oplus\mathbb{Z}_2$, and satisfy only the relation $2a=2b=(2,0)$. Is this the correct way to view it?

Consider the map $f\colon\mathbb{Z}\oplus\mathbb{Z}\longrightarrow\mathbb{Z}\oplus\mathbb{Z}_2$ defined by $f(m,n)=(m,n\pmod2)$. It is a surjective group homomorphism and therefore $\mathbb{Z}\oplus\mathbb{Z}_2\simeq(\mathbb{Z}\oplus\mathbb{Z})/\ker f$. But $\ker f$ is generated by$$(0,2)=2(0,1)=2\bigl((1,1)-(1,0)\bigr)=2(a-b).$$On the other hand, $\mathbb{Z}\oplus\mathbb{Z}=\langle a,b\rangle$. So, $\mathbb{Z}\oplus\mathbb{Z}_2$ is generated by $a$ and $b$ and by the relation $2a=2b$.
$G \cong \langle S | R \rangle$ means that $G$ is the freest group generated by $S$ subject to the set of relations $R$; that is to say, the generators for $G$ are subject to no additional relations not equivalent to the ones already listed.
Your $a = (1,1)$ and $b = (1,0)$ choices generate $\mathbb{Z} \oplus \mathbb{Z}_2$ and satisfy the given relation. To verify that $\mathbb{Z} \oplus \mathbb{Z}_2 \cong \langle a, b \ | \ 2a = 2b \rangle$, we only need to check that $a$ and $b$ are not subject to any additional relations not already implied by $2a = 2b$.
Now, notice that we can rewrite any relation to be in the form $w= \text{id}$ for some word $w$. So, for the problem at hand, when can we have $w = (0,0)$, where $w$ is a word in letters $a$ and $b$? Conveniently, our letters commute, so any such word can be rewritten in the equivalent form $a^nb^m$ for some $n, m \in \mathbb{Z}$. Finally, $a^nb^m = (n+m, \ [m]_2)$. If this is to equal $(0,0)$, we must have $m$ even and $n=-m$; can there be any relations not already implied by $2a = 2b$?