How can I show that two groups represented by permutation are different?

Question

This should be a really basic question, but I can't seem to figure it out explicitly.

How can I show that $$\langle x,y\mid x^8=e, y^2 =e, yxy^{-1} = x^{-1}\rangle \neq \langle x,y\mid x^8=e, y^2 =e, yxy^{-1} = x^{3}\rangle $$

Answer 1

You can observe that the first group (which is just $D_8$) has only two elements of order 4, namely $x^2$ and $x^6$. While, if my computations are correct, the second group has six elements of order 4: $x^2, x^6, xy, x^3y, x^5y, x^7y$.

Usually to prove that two presentations are not isomorphic you should prove that they have different properties: for example one has an element of order $k$ while the other does not, they have different order, they have different numbers of (normal) subgroups, etcetera.

There is also a theorem saying that two finite presentations give isomorphic groups iff you can change one of the presentations to the other via a finite number of precise manipulations of the generators and relations (the Tietze moves). But still this result is difficult to use since it requires a lot of combinatoric computations.

Answer 2

In the first group: $$yx=x^7y$$ $$yxyx=x^7yx^7y=x^{56}y^8=e$$ Therefore $$xyxy=e$$ We have only used that $x^8=e$. For any power $x_0=x^k$ it also holds that, so $$yx^kyx^k=x^kyx^ky=e$$ Now, $$x^kyx^j=x^{k+7j}y^{j+1}$$ So every element of the group if of the form $x^k$ or $x^ky$. All this elements have order $1,2,4$ or $8$. There are $16$ elements: $4$ of order $8$, $2$ of order $4$, $1$ of order $1$ and $9$ of order $2$.

In the second group: $$yx=x^3y$$ $$yxyx=x^3yx^3y=x^{12}y^4=x^4$$ Then $yx$ has order $4$. Since $x^2$ and $x^6$ have also order $4$, both groups are different, because here there are at least three elements of this order... unless $yx=x^2$ or $yx=x^6$. But it if were the case, we'd have $y=x$ or $y=x^5$, but $y$ has order $2$ and $x$ and $x^5$ have order $8$.