I have a question that I feel I am going about in a roundabout way, and would like some help on. I am preparing for an exam.

Problem: Let $G$ be a group with $|G|=150.$ Let $H$ be a non-normal subgroup in $G$ with $|H|=25$.

(a) How many elements of order 5 does $G$ have?

(b) How many elements of order 25 does $G$ have?

My attempt:

$G$ has 6 subgroups of order 25 because the number of subgroups of order 25 has to divide $150/25=6$ and $6 \cong 1 (\mod 5)$. So there are 6 such subgroups: $\{H_1,H_2,H_3,H_4,H_5,H_6\}$. Let $G$ act on this set by conjugation. The permutation representation of the group action of conjugation on this set is $\phi:G \rightarrow S_6$, and $|S_6|=720$. Also, $|\ker(\phi)||im(\phi)|=150$. Because $|im(\phi)|$ has at most one factor of $5$, $5$ divides $|\ker(\phi)|$. Suppose $25$ divides $|\ker(\phi)|$. Then $\ker(\phi)$ has a Sylow 5-subgroup $H_i$. So $H_i \subset \ker(\phi) \implies aH_ia^{-1} \subset a(\ker(\phi))a^{-1} = \ker(\phi)$. By the Second Sylow Theorem, $H_1 \cup \ldots \cup H_6 \subset \ker(\phi) \implies G = \ker(\phi) \implies \phi$ trivial. So $\ker(\phi)$ has a subgroup $K$ with $|K|=5$. So for some $H_i, \ker(\phi) \cap H_i=K \implies a(\ker(\phi))a^{-1} \cap aH_ia^{-1} = aKa^{-1}$

What next? Surely this shouldn't be so long-winded.

  • $\begingroup$ Any ideas anyone? $\endgroup$ – user85362 Dec 6 '13 at 2:46
  • $\begingroup$ Let $K= \ker(\phi)$.Since $6$ divides ${\rm im}(\phi)$, you have $|K| = 5$ or $25$. It can't be $25$ by hypothesis, so $|K|=5$. Now $|G/K|=30$. Show that all groups of order 30 have a normal Sylow 5-subgroup, contradicting your hypothesis. $\endgroup$ – Derek Holt Dec 6 '13 at 8:58
  • $\begingroup$ Could you please clarify why $6$ divides $im(\phi)$ and why K can't be $25$ by hypothesis? $\endgroup$ – user85362 Dec 6 '13 at 13:45
  • $\begingroup$ Can someone please help clarify the comment above? I'm desperately trying to understand this problem in preparation for my final exam. $\endgroup$ – user85362 Dec 6 '13 at 16:02

Trying to clarify Derek Holt argument a bit differently: $H$ acts on the set $Syl_5(G)$ by conjugation, leaving itself invariant. If $L \in Syl_5(G)$, $H \neq L$, then the length of the orbit of $L$ is index$[H:H \cap L] \gt 1$ and divides 5. Since $|Syl_5(G)|=6$ , we conclude index$[H:H \cap L]=5$. So $K=H \cap L$ is a subgroup of order 5 of $H$ and $L$! And this reasoning shows that in fact each pair of Sylow 5-subgroups intersects in $K$. So $K = \bigcap Syl_5(G) \lhd G$ and $|K|=5$. $G/K$ has a normal Sylow 5-subgroup, so $H/K \lhd G/K$, which means after all $H \lhd G$ and hence the exam problem is wrong!

Why does a group $G$ of order 30 always have a normal Sylow 5-subgroup? If it would have 6 (the only other possibility), then the number of elements of order 5 would be 24. That leaves us with 30-24-1=5 elements of order 2 or 3. This implies that a Sylow 3-subgroup $S$ must be normal ($|Syl_3(G)|$ = 1, 2, 5 or 10, and $\equiv 1$ mod 3, but 10 gives us too much elements of order 3!). And $G/S$ of order 10 has definitely a normal Sylow 5-subgroup! If $H$ is a Sylow 5-subgroup of $G$, this means that $HS \lhd G$. But $H$ has index 3 in $HS$, which is the smallest prime dividing $|HS|=15$. So $H$ is normal in $HS$ and in fact characteristic since it is a Sylow subgroup. But then $H \lhd G$, final contradiction.

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    $\begingroup$ No problem, maybe you should try yourself to show that a group of order $pqr$, $p$, $q$, $r$ primes with $p \lt q \lt r$ always has a normal Sylow $r$-subgroup! $\endgroup$ – Nicky Hekster Dec 6 '13 at 23:14

Your first part is correct $G$ has 6 Sylow 5-subgroups. $H$ is abelian and either cyclic or isomorphic to $C_5 \times C_5$. So ...

  • $\begingroup$ Thanks for responding. I don't quite get where you're going, although I'm trying to make use of it. $\endgroup$ – user85362 Dec 5 '13 at 23:42

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