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The cyclic group of order $6$ is the group defined by the generators $a,b$ and relations $a^2=b^3=a^{-1}b^{-1}ab=e.$

For this problem I am assuming that it must be $a,b \neq e$. I will show the $a^ib^j$ are distinct for $i=0,1$ and $j=0,1,2$. If $i=0$, $b^0=e,b,b^2$ are distinct. We know order of $b$ is not $2$ as $b^3=e$ as the order of an element divides powers that make $b$ the identity, this would mean $2\mid 3$ a contradiction. For $i=1$, $$ab^i=ab^k \rightarrow b^i=b^k \rightarrow i\equiv k \mod{3}$$ but as $0\leq i,k < 3$, $i=k$ so $a,ab,ab^2$ are distinct.

So we have that $G \supseteq \{e,a,b,b^2,ab,ab^2\}.$ So $$G=\left<a,b\right> =\{ a^{n_1}b^{n_2}\cdots a^{n_r}b^{n_{r+1}}\mid n_i \in \mathbb{Z}\}$$ and as $a^{-1}b^{-1}ab=e$, $ab=ba$ and so we can write $G$ as $$G=\{a^nb^m \mid n,m \in \mathbb{Z}.$$ We then also have by $a^{-1}b^{-1}ab=e$ that $$a^{-1}b=ba^{-1},\,a^{-1}b^{-1}=b^{-1}a^{-1},\,ab^{-1}=b^{-1}a$$ and so any $$a^nb^m=a\cdots a\cdot b\cdots b =b^ma^n$$ as we can commute the elements to get $b^ma^n$ so $G$ is abelian. To show there are exactly $6$ elements, if we pick any $a^nb^m \in G$, we can assume that $0\leq n\leq 1$ and $0\leq m \leq 2$ and then this element is already accounted for above so $$G=\{e,a,b,b^2,ab,ab^2\}.$$ As $G$ is a finite abelian group of order $6$, $G$ can be written uniquely up to permutation as a direct sum of $\mathbb{Z}_{p_i^{k_i}}$'s where the $p_i$ are prime but there is only one way to do this for $6=2\cdot3$ so this means $$G\cong \mathbb{Z}_2\oplus\mathbb{Z}_3\cong\mathbb{Z}_6$$ as $\gcd(2,3)=1$.

I am just wondering of nicer ways to argue this in the case I would need to do this on an exam. Also if everything looks correct too.

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Your arguments work just fine, but we can do a bit better.

Let's define $$G=\langle a,b\mid a^2=b^3=a^{-1}b^{-1}ab=e\rangle.$$

As you noted, $a^{-1}b^{-1}ab=e$ if and only if $ab=ba.$ Hence, $a$ and $b$ commute, and so since $G$ is generated by $a$ and $b,$ then it follows immediately that $G$ is an abelian group, and in particular $$G=\{a^mb^n\mid m,n\in\Bbb Z\}.$$ Since $a^2=e$ and $b^3=e,$ then we can conclude that $$G=\{e,a,b,ab,b^2,ab^2\}.$$

At this point, there are a few ways we can proceed:

  • Show that $e,a,b,ab,b^2,ab^2$ are distinct, then note that the abelian group of order $6$ is cyclic.
  • Show that $e,a,b,ab,b^2,ab^2$ are distinct, then apply Structure Theorem for Finite Abelian groups, as you did, together with the fact that $\gcd(2,3)=1,$ so that $G$ is cyclic.
  • Show that $e,a,b,ab,b^2,ab^2$ are distinct, then demonstrate an element of $G$ of order $6.$
  • Show that $H:=\langle a\rangle\cong\Bbb Z_2,$ $K:=\langle b\rangle\cong\Bbb Z_3,$ and that $H\cap K=\{e\},$ so since $H,K$ are normal subgroups of $G$ (since $G$ abelian) and $HK:=\{hk\mid h\in H,k\in K\}=G,$ then $G\cong H\times K.$ (Have you seen this result before?) Thus, $G\cong\Bbb Z_2\times\Bbb Z_3\cong\Bbb Z_6.$

That last is arguably the simplest. It suffices to show that $a\ne e,$ $b\ne e,$ $b^2\ne e,$ $a\ne b,$ and $a\ne b^2.$

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