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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.

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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.

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I found $Z(G)=\{1\}$.

Here is my calculation.Let $K=<\sigma>$ and $H=<\tau>$

Since $H$ is cyclic, $H\leq N_G(H)$ and $N_G(H)\cap K=1$ otherwise, $K\leq N_G(H)\implies N_G(H)=G\implies $ $G$ is abelian contradiction.(if $H$ is normal,then $G\cong H\times K$)

Thus,we must have $N_G(H)=H\implies Z(G)\leq H$

Since $K$ is cylic $K\leq C_G(K)$ and since the other elements does not commute with $K\implies$ $ C_G(K)=K \implies Z(G)\leq K$ since $K\cap H=1$ hten we have $Z(G)=1$ .

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  • $\begingroup$ Thank you for this. However, this problem is given before the Sylow theorems in the syllabus, so I guess there should be another solution. $\endgroup$ – Student G Mar 9 '14 at 20:18
  • $\begingroup$ @StudentG: Actuall,it can be done without sylow theorems, let me edit. $\endgroup$ – mesel Mar 9 '14 at 20:21

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