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Problem: Suppose that $G$ is simple group and has an abelian Sylow $2-$subgroup of order $8$. Show that the order of $G$ is divisible by $7$.

Is there any hint to solve this problem? I'll be glad if one gives an answer.

Here's my start:

Let $|G| = 8k$ with $2\nmid k$. Then $n_2 = 1 \mod 2$, and $n\mid k$. Since $G$ is simple, $n_2 \neq 1$. After that I get stuck.

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  • $\begingroup$ what do you think of the question.... ? you tried something? $\endgroup$ – user87543 Jan 12 '14 at 16:50
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    $\begingroup$ Remember that a consequence of Sylow's third is that if there is only one Sylow p-subgroup then it is normal. That will factor into the proof if I'm not mistaken. $\endgroup$ – Nick Jan 12 '14 at 16:55
  • $\begingroup$ Look at the 3 statements in Sylow's theorem. $\endgroup$ – André 3000 Jan 12 '14 at 16:58
  • $\begingroup$ I only do the following: Let $\left| G \right| = 8k$ and $2$ doesn't divide $k$. Then, ${n_2} \equiv 1(2)$ and $\left. {{n_2}} \right|k$. Since $G$ is simple, ${n_2} \ne 1$. How can I show that $\left. 7 \right|k$? $\endgroup$ – egrtomath Jan 12 '14 at 17:16
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    $\begingroup$ Use this fact that $\frac{N_G(H)}{C_G(H)}$ is isomorphic to a subgroup of $Aut(H)$. $\endgroup$ – Babak Miraftab Jan 12 '14 at 17:29
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Isaacs's Prop 5.18 states that whenever $G$ is a finite group with an abelian Sylow $p$-subgroup $P$, then $Z(N_G(P)) \cap G' \cap P = 1$. In our case $G=G'$ so we get that $Z(N_G(P)) \cap P = 1$. Of course $Z(N_G(P)) \cap P = C_P( N_G(P))$ are exactly those elements of $P$ that are left alone by every conjugation from $N_G(P)$.

Since the group of conjugations from $N_G(P)$ is exactly $N_G(P)/C_G(P) \leq \newcommand{\Aut}{\operatorname{Aut}}\Aut(P)$, and since $P \leq C_G(P)$ so that $N_G(P)/C_G(P)$ must be a group of odd order, we are interested in the odd order subgroups of $\Aut(P)$ for $P$ an abelian group of order 8.

If $P=C_8$ then $\Aut(P) \cong C_2 \times C_2$ has no non-identity subgroups of odd order, so $N_G(P)/C_G(P) = 1$ and $N_G(P) = C_G(P)$ and $C_P( N_G(P)) = P \neq 1$. Oops.

If $P=C_4 \times C_2$ then $\Aut(P) \cong D_8$ has no non-identity subgroups of odd order, so oops again.

If $P=C_2 \times C_2 \times C_2$ then $\Aut(P) \cong \operatorname{GL}(3,2)$ has odd order subgroups of orders 1, 3, 7, and 21. The ones of orders 1 and 3 centralize some non-identity elements of $P$, so oops. The ones of orders 7 and 21 are fine.

The one of order 7 creates what is called AGL(1,8) fusion and produces the simple group PSL(2,8). The one of order 21 creates what is called AΓL(1,8) fusion and produces the simple group J1 and ${}^2G_2(3^{2n+1})$ for $n \geq 1$.

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  • $\begingroup$ In the case $P=C_2\times C_2\times C_2$, why elements of orders 1 and 3 centralize some non-identity elements of $P$ and order 7 not? $\endgroup$ – ortmat Jan 13 '14 at 11:27
  • $\begingroup$ You may try listing a few elements of order 1. See if you can find a pattern. $\endgroup$ – Jack Schmidt Jan 13 '14 at 13:05
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Solution: Let $P$ be an abelian Sylow $2−$subgroup of the simple group $G$. If a Sylow $2−$subgroup of a simple group is abelian, then it must be elementary abelian. So, $P \cong {Z_2} \times {Z_2} \times {Z_2}$. Recall that for the elementary abelian group $G$ of order ${p^n}$, $Aut\left( G \right) \cong GL\left( {n,p} \right)$. Note that $\left| {GL\left( {n,p} \right)} \right| = \prod\limits_{i = 0}^{n - 1} {\left( {{p^n} - {p^i}} \right)} $. So, $\left| {Aut\left( P \right)} \right| = 7.6.4 = 168$. Also note that $\frac{{{N_G}\left( P \right)}}{{{C_G}\left( P \right)}} \cong Aut\left( P \right)$ $ \Rightarrow $ $\left| {{N_G}\left( P \right)} \right| = \left| {{C_G}\left( P \right)} \right|.\left| {Aut\left( P \right)} \right| = \left| {{C_G}\left( P \right)} \right|.168$. Hence, $\left. 7 \right|\left| {{N_G}\left( P \right)} \right|$. By Langrange, we are done.

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  • $\begingroup$ Thank you very much Praphulla Koushik and Babgen. $\endgroup$ – egrtomath Jan 12 '14 at 19:21
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    $\begingroup$ In general, you only know that $N_G(P)/C_G(P)$ is isomorphic to a subgroup of ${\rm Aut}(P)$, not necessarily to the whole of ${\rm Aut}(P)$. In fact what you wrote is clearly not right, because $|N_G(P)/C_G(P)|$ must be odd. $\endgroup$ – Derek Holt Jan 12 '14 at 19:26
  • $\begingroup$ @DerekHolt : why should $|N_G(P)/C_G(P)|$ has to be odd :O could you please explain... suppose it is even then it has an element of order $2$ so Automorphism group should have an element of order $2$ ... Is that something which is not quite possible or for some other reason :O Please help me to see this.. $\endgroup$ – user87543 Jan 12 '14 at 19:39
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    $\begingroup$ It has to be odd because $C_G(P)$ contains $P$, which is a Sylow $2$-subgroup of $G$. $\endgroup$ – Derek Holt Jan 12 '14 at 20:24
  • $\begingroup$ @Derek Holt: Thanks a lot. $\endgroup$ – egrtomath Jan 12 '14 at 22:07

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