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Here is a problem from "An introduction to the Theory of Groups" by J.J.Rotman:

Let $G$ be a finite group, and let $H$ be a normal subgroup with $(|H|,[G:H])=1$. Prove that $H$ is the unique such subgroup in $G$.

I assumed there was another normal subgroup like $H$, say $K$, such that $(|K|,[G:K])=1$. My aim was to show that $[K: K\cap H]=1 $ that was not held if I didn’t suppose $|H|=|K|$ . My question is if my last assumption about two subgroups is right? If it isn’t, please guide me. Thanks.

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$[K:K\cap H]=1$ iff $K=K\cap H$ iff $K\le H$. By symmetry, you would also have $[H:H\cap K]=1$ hence $H\le K$ and thus $H=K$. Note that $|H|=|K|$ is not the same as $H=K$ however. –  anon Jul 12 '12 at 14:17
Hint: let $K$ be another subgroup of this order. What can you say about $HK$? –  user27126 Jan 25 '13 at 6:40
Is an assumption missing from the stated claim? Answerers seem to assume that $|H|=|K|$. Why? I do know that the claim is false without that assumption: Let $G=C_6=\langle c\rangle$, $H=\langle c^2\rangle$, $K=\langle c^3\rangle$, and we have $(|H|,[G:H])=1,$ $(|K|,[G:K])=1$?? –  Jyrki Lahtonen Mar 2 '13 at 9:20
@JyrkiLahtonen: I found the number of problem. Do you have the book? It is 2.54. and there is no just a hint "If $K$ be another such subgroup,what happens to $K$ in G/H" –  Babak S. Mar 2 '13 at 10:49
@Babak: I don't have a copy, but the assumption made by the answerers is a must. The exercise makes no sense otherwise. The conditions $H\lhd G, (|H|,[G:H])=1$ are satisfied by all the Sylow subgroups of all nilpotent groups, so there are plenty of counterexamples otherwise. –  Jyrki Lahtonen Mar 2 '13 at 11:45

5 Answers 5

up vote 7 down vote accepted

Alternatively, you might note that if $K$ is any other subgroup of order $|H|$, whether or not $K$ is normal (but assuming $H$ is normal), then $HK$ is a subgroup of $G$, so its order must divide $|G|$.

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This is better, I think. Less technology needed! –  Dylan Moreland Jul 12 '12 at 14:40
@DylanMoreland:You mean we have $\frac{|H||K|}{|H\cap K|}|[G:H]$ and because of our first assumption in the problem, we necessarily get $|H|=|H\cap K|$? –  Babak S. Jul 12 '12 at 15:06
@BabakSorouh [Not sure how these comments got on this answer, sorry Geoff] I guess what I said leads toward saying that $\#(f(K))$ divides $(G : H)$ and also $\#(K) = \#(H)$ because it is a homomorphic image and hence a quotient of $K$. But to be more precise you could restrict the map to $K$ and get $K/(K \cap H) \simeq (KH)/H$, and the order of this last group divides $(G : H)$. Is that what you meant to say? –  Dylan Moreland Jul 12 '12 at 15:19
@DylanMoreland: Yes. Exactly. Thank you and Geoff again. –  Babak S. Jul 12 '12 at 15:24

Let $L$ be any subgroup of $G$ with order $|H|$. Let $p$ be the natural projection $G \to \frac{G}{H}$. Then $p(L)$ is a quotient group of $L$, so $|p(L)|$ divides $|L|=|H|$. But $p(L)$ is also a subgroup of $\frac{G}{H}$, so $|p(L)|$ also divides $|G:H|$. By the coprimality hypothesis, $|p(L)|=1$ so $p$ is trivial on $L$. This means that $L \subseteq H$. Finally $L=H$ because those two subgroups have the same cardinality.

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Very nice answer. –  Ben West Jan 25 '13 at 6:55

One way to see this is to look at the quotient map $f\colon G \to G/H$. Using Lagrange's theorem a couple of times, what can you say about the order of $f(K)$? [This isn't far from proving a well known formula for $\#(HK)$.]

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Consider the more general result proven by user Gastón Burrull here. It states

If $K\triangleleft G$, $|K|$ finite, $H\leq G$, $[G:H]$ finite and $|K|$, $[G:H]$ are relatively prime then $H\leq K$

In other words, if the index of a subgroup is coprime to the cardinality of a normal subgroup, then it is contained in that normal subgroup. So if $K$ is any other subgroup of cardinality $|H|$, then since the group is finite, $H$ and $K$ have the same index, hence the index of $K$ is coprime to $|H|$, and so $K\leq H$. But since $K$ and $H$ are finite subgroups of equal order, $K=H$. So $H$ is the only subgroup of order $|H|$.

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This machinery is somewhat high powered, but it should explain what the book is really trying to get at.

Hall subgroups are generalizations of Sylow subgroups for multiple primes. If we denote by $\nu_G(p)$ the largest power of $p$ dividing $|G|$, we see that Sylow $p$-subgroups are those subgroups of order $\nu_G(p)$. Hall $\pi$-subgroups, where $\pi$ is a set of prime numbers, are subgroups of order $\prod_{p\in \Pi} \nu_G(p)$. Alternatively, a Hall $\pi$-subgroup $H$ is a subgroup of order divisible by each $p\in \Pi$ such that $[G:H]$ is coprime to $|H|$. Note that Hall subgroups of a group $G$ may not always exist for each $\pi\subset\{p\in \mathbb{P}|p\mid |G|\}$ - actually, this holds if and only if $G$ is solvable.

As it turns out, a lot of the stuff that works for Sylow subgroups works for Hall subgroups. In particular, when Hall subgroups exist, they are all conjugate, so much like how normal Sylow subgroups must be unique, so must normal Hall subgroups. This is part of Hall's theorem, which I will not prove here (but it follows fairly easily by inducting on the size of a minimal normal subgroup).

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Thanks for the time you gave me Alexander. I appreciate you. –  Babak S. Apr 11 '13 at 7:48

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