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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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You should be able to prove that using the fact that the multiplicative group of a finite group is cyclic. The elements of order dividing $q-1$ in a cyclic group of order $q^2-1$ are all squares. – Derek Holt Jun 1 '12 at 19:18
Alternately, use the fact that the field of order $q^2$ is unique. – Qiaochu Yuan Jun 1 '12 at 19:20
@QiaochuYuan: You meant that, according to Derek's note, all elments of order $q+1$ in $GF(q^2)$ make a field of order $q$? – Babak S. Jun 1 '12 at 19:28
I mean that if $r \in \mathbb{F}_q$, either the polynomial $x^2 - r$ is reducible over $\mathbb{F}_q$ (in which case $r$ already has a square root in $\mathbb{F}_q$) or it is irreducible over $\mathbb{F}_q$ (in which case its splitting field is a quadratic extension of $\mathbb{F}_q$, which is unique: $\mathbb{F}_{q^2}$). – Qiaochu Yuan Jun 1 '12 at 19:31
@QiaochuYuan: Thanks. You two made me lots of ideas here. – Babak S. Jun 1 '12 at 19:34
up vote 3 down vote accepted

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.

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