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$G$ is a finite group generated by two elements $a$ and $b$, we are given the following data:

Order of a= $2$

Order of $b=2$

Order of $ab=8$.

If $Z(G)$ denotes the center then what is $G/Z(G)$ isomorphic to?

Attempt:

To be honest I don't know how to start this. I thought of taking $D_8$ as a concrete example of such a group but was not able to proceed much.

One thing I can see that this group is non abelian as if it was abelian then it would imply that $(ab)^2=e$ which contradicts the fact that order of $ab$ is 8.

What can I say more about this group?

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$G$ is indeed the dihedral group of order $16$.

A useful fact: The center of dihedral group $D_{2n}$ (notation: $D_{2n}$ is the dihedral group of order $2n$) is trivial if $n$ is odd and is $\pm 1$ if $n$ even. This is easily seen from the relation $r\bar{r}=\bar{r}r^{-1}$.

For $n>2$, the quotient $D_{2n}/Z$ is also generated by two elements of order $2$, so is dihedral of the appropriate order.

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  • $\begingroup$ The quotient doesn't appear to be dihedral. $\endgroup$ – Chris Custer Jan 3 at 3:37
  • $\begingroup$ @ChrisCuster Huh? It is dihedral from the relations (which automatically hold for quotients) $\bar{r}^2=(r\bar{r})^2=1$ and order. $\endgroup$ – user10354138 Jan 4 at 22:47
  • $\begingroup$ Oops. My mistake. $\endgroup$ – Chris Custer Jan 5 at 0:48
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In general, $G/Z(G)\cong\operatorname{Inn}G$, the group of inner automorphisms of $G$.

Since in this case $G=D_{16}$ (see this), and $Z(D_{16})=\{1,r^4\}$, we have $\operatorname{Inn}G\cong D_{16}/\mathbb Z_2\cong D_8$.

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