Find order of group given by generators and relations Let $G$ be the group defined by these relations on the generators $a$ and $b$: $\langle a, b; a^5, b^4, ab=ba^{-1}\rangle$. I need hints how to find order of $G$.
 A: Since $G$ is generated by $a,b$, the general element is of the form $$\tag1a^{n_1}b^{m_1}a^{n_2}b^{m_2}\cdots a^{n_k}b^{m_k}$$ with $n_i, m_i\in\mathbb Z$.
However, $a^5=1$ implies that we can restrict to $0\le n_i\le 4$ and $b^4=1$ implies that we can restrict to $0\le m_i\le 3$.
The last relation (rewritten as $ab=ba^3$ by our wish to have $0\le n_i\le 4$) shows that we can "pull" all $b$s to the left of all $a$s, but also the other way: $$\tag 2a^2b=ba^6=ba$$ allows us to pull $a$s to the left instead.
Now among all representations of an element $g\in G$ we may pick one that minimizes $k$ in $(1)$. Use the relation $(2)$ to arrive at a contradiction against the minimality of $k$ if $k>1$.
So far we have shown that $|G|\le 20$. You could in principle write down a multiplication table for these twenty elements $a^nb^m$. Alternatively, exhibit a well-known group of order $20$ that has generators obeying the given relations.
To the latter end, we may want to rewrite $ab=ba^{-1}$ as $b^{-1}ab=a^{-1}$. That is: The cyclic group generated by $b$ (which we expect to be isomorphic to $\mathbb Z/4\mathbb Z$) acts on the cyclic group generated by $a$ (which we expect to be isomorphic to $\mathbb Z/4\mathbb Z$) by conjugation. This makes $\langle a\rangle$ a normal subgroup of $G$. Moreover, the action by the generator $b$ is "take the inverse" (indeed an automorphism of $\mathbb Z/5\mathbb Z$ of order dividing $4$). Thus we identify $G$ as semidirect product $\mathbb Z/5\mathbb Z\rtimes\mathbb Z/4\mathbb Z$.
A: The only good way to do problems like this is to play around with words in the letters $a,b,a^{-1},b^{-1}$, seeing if you can use those given relations to get any word in some standard form. Then, you can say every element in $G$ is equivalent to one in this form, and there are so many words in that form, so $G$ has this order.
In your case, the $ab=ba^{-1}$ relation is powerful: it allows you to essentially move $a$'s though $b$'s (though it inverts the $a$'s in the process), so you can group all the $a$'s and $b$'s in any word together. Then you can use the $a^5$ and $b^4$ relation to reduce the size of the $a$ and $b$ clumps.
A: We can show by induction that any word in $a, b$ (and their inverses) is equivalent under the relators to a unique word of the form $$a^k b^l$$ where $0 \leq k <5 $ and $0 \leq l < 4$.
