I am studying for a qualifying exam and am working on the following problem.

If G is a group of order 2010=2*3*5*67, then G has a normal subgroup of order 5.

I know that the number of 67-Sylow subgroups is 1, lets's call it H. So H is normal in G. We also know that the order of H is 67.

Since sylow-p subgroups exist, there is a subgroup K with order 5.

Since H is normal and K is a subgroup, HK is a subgroup of G.

Since the orders of H and K are relatively prime, their intersection is trivial so the order of HK is 5*67.

Also since the orders of H and K are prime, they are cyclic, and again because they are relatively prime, HK is cyclic and hence abelian.

Since abelian groups have a subgroup the size of each divisors of its order, then HK has a subgroup of order 5, call it M.

This is where I am stuck. My study buddy says HK is normal, but can't give me a reason.
He states exactly that "M is a subgroup of HK, HK is normal in G, and since HK is cyclic, we can conclude that M is normal in G."

I don't see why HK is normal and how you can conclude that M is normal in G because HK is cyclic.

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    $\begingroup$ If you're taking the qual in the spring, you should be okay, since 2017 is prime. $\endgroup$ – Tim kinsella Aug 21 '16 at 3:50
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    $\begingroup$ A couple small notes: $HK$ is cyclic because 5 doesn't divide $67-1$, not because 5 and 67 are relatively prime. And you can take $M$ to be $K$ if you like--in fact, once you've solved the problem, you will know they're equal. $\endgroup$ – Ravi Fernando Aug 21 '16 at 4:26
  • $\begingroup$ I see what you mean about HK being cyclic because 5 does not divide 67-1. I just found that theorem about groups of order a product of two primes where p < q and p not dividing q-1 implies the group is cyclic. It makes sense that you say M can be taken to be K since we know we have to prove it is normal so there can be only one sylow-5 subgroup. How can I get to HK being normal? $\endgroup$ – Amanda Aug 21 '16 at 4:40

If $HK$ is a cyclic normal subgroup of $G$, then any subgroup of $HK$ is characteristic in $HK$, and therefore normal in $G$ by this. It's not immediately clear to me how to prove this directly, but here's another approach.

First, there's a general fact that any group of order 2 mod 4 has a subgroup of index 2. Proof: embed $G$ in $S_{|G|}$ by the left-multiplication action, and intersect with $A_{|G|}$. This will have index 2 in $G$ unless $G$ is already contained in $A_{|G|}$. But it isn't: any order-2 element in $G$ acts as $|G|/2$ transpositions, which is an odd permutation.

So $G$ must contain a subgroup $L$ of order 1005, necessarily normal in $G$. If we can show that $L$ has a characteristic subgroup of order 5, then the link above proves that this subgroup is normal in $G$. To accomplish this, we can go back to Sylow's theorem, and use some of the ideas you already came up with for $G$.

First, Sylow's theorem tells us that the number of Sylow 5-subgroups of $L$ is either 1 or 201. If there is only one, then it's characteristic, so suppose for contradiction that there are 201. This would account for 804 nonidentity elements, which leaves only 201 elements of $L$ not of order 5. But by your argument, there is a unique Sylow 67-subgroup of $L$, and its product with any order-5 subgroup is a cyclic subgroup of order 335. This must have $\varphi(335) = 4 \cdot 66 = 264$ generators, which is too many. So $L$ must have a single Sylow 5-subgroup, which is therefore characteristic in $L$ and thus normal in $G$.

  • $\begingroup$ +1, nice answer. The argument for the normality of the subgroup $L$ of order $5$ was very elegant. $\endgroup$ – Alex Wertheim Aug 21 '16 at 5:13
  • $\begingroup$ This looks great and I like that its tied to the theorem regarding being characteristic. I remember proving that a while back. $\endgroup$ – Amanda Aug 21 '16 at 5:48
  • $\begingroup$ Can you explain the part about, this must have φ(335)=4⋅66=264φ(335)=4⋅66=264 generators a little more please? $\endgroup$ – Amanda Aug 21 '16 at 5:49
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    $\begingroup$ @Amanda: each of the $264$ generators of the cyclic group of order $335$ have order $335$. If there are 804 elements of order $5$ in $L$, then together these account for $804+264 = 1068 > 1005$ elements of $L$, a contradiction. If you're asking why there are $264$ generators, it's because a residue class of $C := \mathbb{Z}/335\mathbb{Z}$ generates $C$ if and only if its representative is relatively prime to $335$. There are therefore $\varphi(335) = \varphi(5)\cdot\varphi(67) = 4 \cdot 66$ such elements. $\endgroup$ – Alex Wertheim Aug 21 '16 at 6:08
  • $\begingroup$ That was very helpful, thanks so much. $\endgroup$ – Amanda Aug 21 '16 at 7:05

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