Find the center of a specific group The group $G$ is generated by the two elements $\sigma$ and $\tau$, of order $5$ and $4$ respectively. We assume that $\tau\sigma\tau^{-1}=\sigma^2$.
I have shown the following:
* $\tau\sigma^k\tau^{-1}=\sigma^{2k}$ and $\tau^k\sigma\tau^{-k}=\sigma^{2^k}$.
* $\langle\sigma\rangle$ is a normal subgroup of $G$, and $\langle\sigma\rangle\cap\langle\tau\rangle=\{e\}$.
* $G/\langle\sigma\rangle=\langle\tau\langle\sigma\rangle\rangle$.
* $G$ is of order $20$ and every element $g$ in $G$ may be written uniquely in the form $g=\sigma^k\tau^m$, where $0\le k<5$ and $0\le m<4$.
* The commutator subgroup $[G:G]=\langle\sigma\rangle$.
What remains is to find the center $Z(G)$ of $G$. Any suggestions on how to proceed? Thank you.

You have a normal subgroup of order $5$. Your calculations are already sufficient to show that the elements of this group other than the identity don't commute with $\tau$, or indeed any of its powers. So $\tau, \tau^2, \tau^3$ are not in the centre. $1=\tau^0=\tau^4$ is of course in the centre.
Suppose we have an element $\rho$ which is in the centre, and therefore does commute with $\tau$. You have shown that $\rho = \sigma^k\tau^m$ so that $$\rho \tau =\sigma^k\tau^{m+1}$$
$$\tau\rho=\tau\sigma^k\tau^m=\sigma^{2k}\tau^{m+1}$$
For these to be equal you need $k=0$ - so the only possible central elements other than $1$ are the powers of $\tau$ which have already been excluded.
