Order of a subgroup generated by two elements in $S_5$ Let $G = \langle(12)(34), (15)\rangle$ be a subgroup of $S_5$.
Then I need to show that $G$ has order $12$ and has a non-trivial centre.
I have found thse elements- $$I,(12)(34), (15), (12)(34)(15), (15)(12)(34).$$
If I just keep computing compositions, then the whole process is becoming really cumbersome.
Please, help!
 A: This $G$ is a group generated by two involutions. So it is a dihedral group $D_n$ for some $n$. This  $n$ is the order of the product of the involutions $(15)(12)(34)=(152)(34)$ which is $6$, so your group $G$ is isomorphic to $D_6$ and has 12 elements.
A: Let $a = (12)(34)$ and $b = (15)$. Clearly $a^2 = b^2 = e$, where $e$ is the identity.
Define $u = ab$. It is easy to check that $u$ has order six, and also that $auau = e$. I can add detail if you need me to, but this seems like a good exercise to me.
The dihedral group of order $12$ (the symmetries of a regular hexagon) has presentation
$$
D_{12} = \langle \sigma, \tau \mid \sigma^6 = \tau^2 = \sigma \tau\sigma\tau = e \rangle
$$
We have verified that $a, u$ satisfy the defining relations of $D_{12}$, which means that the group generated by $a$ and $u$ has size at most $12$ (intuitively, this is because any combination of them that equals the identity in $D_{12}$ will also equal the identity in $G$, so there can be at most twelve distinct combinations. This follows very easily from some slightly more advanced theory but I'll try to avoid that).
Since $b = a^{-1}u$, the group generated by $a$ and $u$ is actually, $G$, so $G$ has size at most twelve. But $u$ is an element of order six, so the order of $G$ is a multiple of six. The cyclic group of order six has only one element of order six, but $a$ and $b$ are distinct elements of order $2$, which means that $G \neq \langle u \rangle \cong C_6$, and hence $\lvert G \rvert > 6$, so $\lvert G \rvert = 12$.
A: One way is to use GAP, like so:
gap> S:=SymmetricGroup(5);
Sym( [ 1 .. 5 ] )
gap> G:=Subgroup(S, [(1,2)*(3,4), (1,5)]);
Group([ (1,2)(3,4), (1,5) ])
gap> Size(G);
12
gap> Centre(G);
Group([ (3,4) ])
gap> Size(last);
2
gap>

