are these two finite groups with different presentation isomorphic?

Question

Consider two groups $$\langle x,y \, | \, x^4=y^5=1 ,yxy=x \rangle$$ and $$ \langle a,b \, | \, a^{10}=1,b^2=a^5,aba=b \rangle.$$

I think they are isomorphic, but I can't show it, it will be great if you help about this.

I have the software GAP, if you tell how can I show it with GAP, it will be enough.

thanks.

I did big mistake when I was typing the second group, I edit it now, please excuse me.

actually second group is $Dic_4$,Dicyclic group of order 20.

Answer 1

Since there was a question of how to do it in GAP:

You first create the two groups:

f:=FreeGroup("x","y");
rels:=ParseRelators(f,"x4,y5,yxy=x");
g1:=f/rels;
f2:=FreeGroup("a","b");
rels2:=ParseRelators(f2,"a10,b2=a5,aba=b");
g2:=f2/rels2;

Then (this implicitly uses that the groups are finite, it will not necessarily work for infinite groups):

gap> IsomorphismGroups(g1,g2);
[ x, y ] -> [ b*a^31, a^24 ]

returns one (not necessarily the nicest, or always the same) isomorphism (or fail if none exists). If you want to try all possibilities, you could use

SetReducedMultiplication(g2);
SetReducedMultiplication(g1);

to force immediate reduction of word expressions (otherwise GAP builds up products and never reduces) and then use (to get all isomorphisms, multiply one particular one with all possible automorphisms of the image):

a:=AutomorphismGroup(g2);
all:=List(Elements(a),x->iso*x);

and find (this is for me the nicest) the isomorphism $x\mapsto b$, $y\mapsto a^2$.

Answer 2

Let $G=\langle x,y \ | \ x^4=y^5=1, \ yxy=x\rangle$ and $H=\langle a,b \ | \ a^{10}=1, b^2=a^5, aba=b\rangle.$ Note that $x^2\in Z(G)$.

Define $\phi\colon H\to G$ by \begin{align} \phi(a) &= x^2 y, \\ \phi(b) &= x. \end{align} Then \begin{align} \phi(a^5)&=(x^2 y)^5=x^{10}y^5=x^2=\phi(b^2),\\ \phi(aba)&=(x^2 y)x (x^2 y) = yxy = x = \phi(b),\\ \phi(b^2 a)&=y. \end{align} Thus $\phi$ is a well-defined surjective homomorphism of groups.

We need to verify injectivity. Note that if we knew that $G$ has order 20, then we are done, but by OP we only know that $H$ is order 20; surjectivity forces $G$ finite, at least. So we either verify that $G$ has order (at least) 20, or verify that $\ker(\phi)$ is trivial. I'll do the latter, since then we don't even need to know or check that either group is finite.

For any integers $i,j$ we have $$ \phi(a^i b^j) = x^{2i}y^i x^j = x^{2i+j} y^{(-1)^j i}.$$ So for this to be the identity we have $i\equiv 0\bmod 5$ and $2i+j\equiv 0\bmod 4$. So we may write $i=5k$ for $k=0,1$ and then $10k+j\equiv 2k+j\equiv 0\bmod 4$. Solving, we get the only solutions with $0\leq j<4$ are $k=0, j=0$ and $k=1,j=2$. In both cases, $a^i b^j = a^{5k}b^j=a^{10} = 1$. Thus $\ker(\phi)=1$ and the homomorphism is injective.

Thus we have exhibited a group isomorphism, and $G\cong H$.