Find the group with presentation $\langle x, y: xy=yx, x^5=y^3\rangle$. Here's the problem:

Find the group with presentation $\langle x, y: xy=yx, x^5=y^3\rangle$.

Here's my answer:

The group is commutative by the first relation, so every "word" is of the form $x^iy^j$ for $i,j \in \mathbb{Z}$. By the second relation we have $x^a y^b $ for $0 \leq a \leq 4$ and $b \in \mathbb{Z}$. Hence the group is  isomorphic to $\mathbb{Z_5} \times \mathbb{Z}$.

Is this correct?
Your answer can't be right, since multiplying any element in $ \mathbb Z/5\mathbb Z \times \mathbb Z $ by $ 5 $ gives you something of the form $ (0, 5x) $, and such an element can't be divided by $ 3 $ in this group, so there's no element to play the role of $ y $.

To find the isomorphism class, note that there is a surjective map $ \mathbb Z^2 \to G $ given by $ (n, m) \to x^n y^m $  (where $ G $ denotes your group) whose kernel is precisely $ \langle (5, -3) \rangle $. In other words, we're trying to find the quotient $ \mathbb Z^2/\langle (5, -3) \rangle $.

This quotient is isomorphic to $ \mathbb Z $, as you can see by noting that $ (5, -3) $ and $ (3, -2) $ form a basis of $ \mathbb Z^2 $, so $ G \cong \mathbb Z $.
Is this correct?


No.

gap&gt; F:=FreeGroup(2);
&lt;free group on the generators [ f1, f2 ]&gt;
gap&gt; rels:=[(F.1)*(F.2)*((F.1)^(-1))*((F.2)^(-1)), (F.1)^5*((F.2)^(-3))];
[ f1*f2*f1^-1*f2^-1, f1^5*f2^-3 ]
gap&gt; G:=F/rels;
&lt;fp group on the generators [ f1, f2 ]&gt;
gap&gt; P:=PresentationFpGroup(G);
&lt;presentation with 2 gens and 2 rels of total length 12&gt;
gap&gt; TzGoGo(P);
#I  there are 2 generators and 2 relators of total length 12
gap&gt; TzPrint(P);
#I  generators: [ f1, f2 ]
#I  relators:
#I  1.  4  [ 1, 2, -1, -2 ]
#I  2.  8  [ -2, 1, 1, 1, 1, 1, -2, -2 ]
gap&gt; Size(G);
infinity
gap&gt; StructureDescription(G);
"C0"
gap&gt;