notations of generators and relations

Question

I need to understand the following about generators and relations notations:

• Is $\langle a,b \mid a^kb^l\rangle =\langle a,b\mid a^k=b^l\rangle =\langle a,b\mid a^k,b^l\rangle$?

• Is $ \langle a,b\mid a^k=b^l\rangle =\langle a,b\mid a^k=b^l=1\rangle$? (if not, why?)

• Which of them is $\mathbb Z_k \ast \mathbb Z_l$?

Answer 1

In a presentation of a group the relations are understood to be equal to the identity. So in your examples

$\langle a, b| a^kb^l\rangle$ is generated by $a$ and $b$ and $a^kb^l=e\Rightarrow a^k=b^{-l}$. From this we can see that the three presentations in the top row are not the same. Note however that even though the presentations are not the same does not mean that the groups are not isomorphic.

In particular, $\langle a, b| a^kb^l\rangle\neq\langle a, b| a^kb^{-l}\rangle=\langle a, b| a^k=b^l\rangle$, where the $\neq$ means that presentations are not the same, but the groups are isomorphic.

With regards to your final question

$\mathbb{Z}_k\ast\mathbb{Z}_l\simeq\langle a, b| a^k, b^l\rangle$

Answer 2

We have $\langle X|R\rangle=\langle X|S\rangle$ iff the relations $R$ and $S$ are logically equivalent. In particular, if they are logically equivalent, then they are logically equivalent for arbitrary groups. Thus to disprove equality it suffices to exhibit groups with elements (identified with $X$) satisfying relations $R$ that do not satisfy the relations $S$, or vice-versa.

Here we are thinking of equality as stronger than isomorphism. By $\langle X,R\rangle=\langle X,S\rangle$ we mean that not only are the groups isomorphic, but that $x\mapsto x$ ($x\in X$) extends to an isomorphism.

For example, to disprove $\langle a,b|a^kb^l\rangle=\langle a,b|a^k=b^l\rangle$, it suffices to exhibit a group $G$ with $a,b\in G$ satisfying $a^kb^l=e$ but not satisfying $a^k=b^l$. The idea is that the only way $a^k=b^l$ is compatible with $a^kb^l=e$ ($\Leftrightarrow a^k=b^{-l}$) is if $a^k$ has order $\mid2$ ($x=x^{-1}\Leftrightarrow x^2=e$), so let's force it to not have order dividing $2$ but still have some $l$th root. In particular, let $a=l$, $b=-k$ within ${\bf Z}/3kl{\bf Z}$. Then $a^k$ written additively is $kl$ and $b^l$ is $-kl$, so $a^kb^l=e$. But $a^k\ne b^l$ as $kl\ne- kl$ mod $3kl$.

Now you try.

For the free product, which "glues" groups together, we have $\langle X,R\rangle*\langle Y,S\rangle\cong\langle X\cup Y|R\cup S\rangle$ when $X$ and $Y$ are disjoint. Thus $C_k*C_l=\langle a|a^k\rangle*\langle b|b^l\rangle\cong\langle a,b|a^k,b^l\rangle=\langle a,b,a^k=b^l=e\rangle$.