Finite groups with the same presentation are isomorphic, has the 'same' center and conjugacy classes?

Question

I was considering the following group:

$G = \langle(12),(1425)\rangle \leqslant S_5.$

If we call $s = (12)$ and $r=(1425)$, obviously we have $r⁴=1=s²$.

Also, we can see that $srs^{-1} = (2415) = (1524) = r^{-1}.$

So $G$ satisfies the same presentation as $D_4 = \langle x,y:x⁴=y²=1,yxy^{-1}=x^{-1}\rangle.$

Those relations must give us $G = \{1,r,\cdots,r³,s,\cdots,r³s\}$ and similarly for $D_4$. Hence we can construct an explict isomorphism given by $\phi : \; G \rightarrow D_4$ by $r^is^j\mapsto x^i y^j$.

Also, it follows that the centers are isomorphic. It gives us that $Z(G)= \{1,r²\}, $ since we have $Z(D_4)$ = {1,x²}.

Also, by calculations I could see that the conjugacy classes for $G$ are:

$\{1\}, \{r²\}, \{r,r³\},\{s,r²s\},\{rs,r³s\}$, which are the same as $D_4$.

So,

my questions are: Is this always true in general?

Two groups with the same presentation are always isomorphic? do they have the same center and conjugacy classes?

Thanks in advance!

Answer 1

So $G$ satisfies the same presentation as $D_4$

Be careful! You've shown that it has two generators ($r$ and $s$), and that those generators satisfy a bunch of relations ($r^4 = s^2 = 1$ and $srs^{-1} = r^{-1}$). This is enough to show that $G$ is isomorphic to some quotient of $D_4$. (Use the first isomorphism theorem: map $x$ to $r$ and $y$ to $s$, and show that this map gives a surjective homomorphism.)

But there's one thing you haven't checked: are you sure $r$ and $s$ don't satisfy any extra relations? (Equivalently: what's the kernel of this surjective homomorphism?)

It turns out that they don't, by the way (i.e. this homomorphism is also injective). But you need to check that.

Two groups with the same presentation are always isomorphic?

Yes.

do they have the same center and conjugacy classes?

Isomorphic groups have the same everything - except that their elements might be called different things.