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Let $p,q$ be primes such that $p$ is a divisor of $|\text{SL}_2(q)|=(q-1)q(q+1)$. Hence $\text{SL}_2(q)$ admits non-trivial Sylow subgroups. I am interested in the isomorphism type of the $p$-Sylow. From what I understand, if $p<q$ are odd primes, then a $p$-Sylow subgroup of $\text{SL}_2(q)$ is cyclic.

However, in the case $p=2$ or $p=q$, I am not sure.

I tried to use Suzuki paper(http://www.jstor.org/stable/2372591?seq=1#page_scan_tab_contents), but it is rather complicated and I hoped there is an easier approach.

Thanks in advance for any help.

Ofir Schnabel
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2 Answers2

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${\rm SL}(2,q)$ has a unique element of order $2$, namely $-I_2$. So the same applies to its Sylow $2$-subgroups. There is theorem that the only $2$-groups with this property are cyclic and generalized quaternion. Since the Sylow $2$-subgroups are not cyclic, they are geenralized quaternion.

Derek Holt
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  • The last statement is unclear to me. (I mean: if you write Since Sylow-2 subgroups are not generalized quaternion, they are cyclic, then also I didn't see why it is not true. In short, the reasoning is unclear to me.) – p Groups Jan 27 '16 at 03:14
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We know that the group $SL_2(3)$ has $Q_8$ as $2$-Sylow-subgroup, which certainly is not a cyclic group - see here. This is the case $p=2$, $q=3$, satisfying $p\mid (q-1)q(q+1)$.

Edit: For odd primes $p$, every Sylow $p$-subgroup of $SL(2,q)$ is indeed cyclic (see here, G6f3 c)).

Dietrich Burde
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  • I said it was cyclic when both $p,q$ are odd primes. – Ofir Schnabel Jan 26 '16 at 12:08
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    You also said "However, in the case $p=2$ I am not sure". Hence my example. – Dietrich Burde Jan 26 '16 at 12:10
  • OK, sorry and thanks for your example. Do you know what happens in the case where $p=q$ is an odd prime? – Ofir Schnabel Jan 26 '16 at 12:11
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    The group $SL_2(q)$ clearly has a cyclic subgroup of order $q-1$ as the subgroup of diagonal matrices, and a cyclic subgroup of order $q+1$ that we get as the intersection of $SL_2(q)$ and the cyclic group $\Bbb{F}_{q^2}^*$ of the quadratic extension field (viewed as a subgroup of $GL_2(q)$. Therefore for $p>2$ a Sylow $p$-subgroup is cyclic. Even if a higher power of $p$ happens to be a factor of either $q-1$ or $q+1$. – Jyrki Lahtonen Jan 26 '16 at 14:05
  • @JyrkiLahtonen thanks. But is it also true in the case $p=q$ is an odd prime? – Ofir Schnabel Jan 26 '16 at 14:13
  • @OfirSchnabel The Sylow $q$-subgroups of $SL_2(q)$ are cyclic and conjugate to the subgroup of upper unit-triangular matrices: $\begin{bmatrix}1&x\0&1\end{bmatrix}$. – David Hill Jan 26 '16 at 17:12
  • @DavidHill Thanks a lot. – Ofir Schnabel Jan 27 '16 at 07:37