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I am just reading the proof of a theorem about

$M(q^2)$=$PSL_2(q^2)\bigcup \left\{f|f(z)=\frac{az^q+b}{cz^q+d}, 0≠ad-bc≠k^2; a,b,c,d\in GF(q^2); z\in GF(q^2)\cup\left\{\infty\right\} \right \}$

telling that:

The groups $M(q^2)$ and $PGL_2(q^2)$, wherein $q$ is an odd number, are not isomorphic.

Somewhere in the middle of the proof, it is assumed that the elements of $GF(q)$ are squares in $GF(q^2)$. Any sparks for solving this step of the Theorem. Thanks.

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    Qiaochu's hint is **the** way. This can be done in several ways, but... How do you turn an element of a field to a square? You adjoin its square root! Done.2012-06-01

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Let $\,\mathbb{F}_n\,$ be the field with $\,n\,$ elements. Define $\,\,f:\mathbb{F}_{q^2}^*\to \mathbb{F}_{q^2}^*\,\,$ by $\,\,f(r):= r^2\,\,$ and check its kernel $\ker f:= \{r\in\mathbb{F}_{q^2}^*\,\,;\,\,r^2=1\}$

By the first isomorphism theorem for groups get that $q^2-1=\left|\mathbb{F}_{q^2}^*\right|=2\,\left|f\left(\mathbb{F}_q^*\right)\right|\Longrightarrow\left|f\left(\mathbb{F}_q^*\right)\right|=\frac{q+1}{2}(q-1)$

Thus the subgroup of square elements in $\,\mathbb{F}_{q^2}^*\,$ is divisible by $\,q-1\,$ . Apply now Derek's hint above.