4
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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 $2a=2b$.

$\endgroup$
2
$\begingroup$

Good work so far; you are on the right track.

$G \cong \langle S | R \rangle$ 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 $2a = 2b$?


Footnote: Apologies for any confusion arising in the last paragraph due to my switch between multiplicative notation for general groups and the standard additive notation for the integers and their quotient groups.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.