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 $\mathbb Z_k \ast \mathbb Z_l$?


2 Answers 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 $C_k*C_l=\langle a|a^k\rangle*\langle b|b^l\rangle\cong\langle a,b|a^k,b^l\rangle=\langle a,b,a^k=b^l=e\rangle$.

  • $\begingroup$ Thank you very much! this is helpful $\endgroup$
    – Ronald
    Jul 16, 2013 at 4:48

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

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

  • $\begingroup$ Thank you, this is nice explanation, so are all of the presentations in the question isomorphic to $\mathbb{Z}_k\ast\mathbb{Z}_l$? $\endgroup$
    – Ronald
    Jul 16, 2013 at 4:37
  • 2
    $\begingroup$ No. $\langle a, b| a^kb^l\rangle$ is not. $\endgroup$ Jul 16, 2013 at 4:43
  • $\begingroup$ It is worth pointing out that $\langle X; u=w\rangle$ is not always isomorphic to $\langle X; uw\rangle$. For example, take $X=\{a\}$ and $u=a=w$. One defines the infinite cyclic group, while the other defines the cyclic group of order two. The key "bit" is that here, $u$ and $w$ consist of words in different generators, and that there exists some automorphism of $F(X)$ with $w\mapsto w^{-1}$. These are pretty specific conditions. $\endgroup$
    – user1729
    Jul 16, 2013 at 8:25

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