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This time I'm having trouble with the following exercise:

Being $|G| = 20$ and $H$ and $K$ subgroups of $G$ whose order is 5, prove that $K = H$.

I'm also recommended to start by proving that $|H \cap K| = 5$.


My draft:

Let $x \in (H \cap K)\setminus{\{e\}}$. Then $|x|$ divides $|H| = |K| = 5$. Because $x \neq e$, $|x|$ can only be 5. So $| = H \cap K| = 5$.

Is this part OK?

But now, on to prove that $H = K$, I've been wandering around unsuccessfully (my rubber is already half the size it was earlier today :p). Can you drop any hint on this matter?

Thanks for taking the time to read!

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    Do you know Sylow's theorems?2011-02-18
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    @Chris Eagle: No. This exercise is presented after Lagrange's theorem.2011-02-18
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    On a sidenote, the notation $|H\leq G| = 5$ isn't quite correct I think (I've never seen it anyway), if what you mean is actually $|H|=5$ and $H\leq G$.2011-02-18
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    @Myself: That's exactly what I meant. I chose it so to be brief, but perhaps it's not adequate. My apologies.2011-02-18
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    @Marla: Yes, I'm afraid that notation is incoherent. You should edit the title...2011-02-18
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    @Arturo Magidin, @Myself: Done editing the title. Hope it's fine now.2011-02-18
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    @Marla: Once you show that $H\cap K$ has five elements in it, you're done: $H\cap K$ is contained in $H$, and also contained in $K$. And $H\cap K$ has five elements, and $H$ has five elements, and $K$ has five elements...2011-02-18
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    An alternative way to do this is to look at the set HK and note that it must have size 5 since it is clearly contained in G.2011-02-18
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    @Arturo Magidin: How I overlooked that?! That's it! Thank you!2011-02-18

2 Answers 2

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I'm afraid your draft isn't quite complete, because you're assuming such $x$ exists. In other words, you're assuming $H\cap K$ is non-trivial. However, if you could exlude that case then you would be done.

Here's a complete reasoning, don't read it if you want to keep looking :-)

You should note that $H\cap K\leq H$, therefore $ |K\cap H|\mid |H| = 5$, because of Lagrange's theorem. In other words $|H\cap K| = 1$ or $|H\cap K| = 5$, because $5$ is prime.

Now assume $|H\cap K| = 1$ indeed, (we will try to exclude this case, by deriving a contradiction). Then $|HK| = \frac{|H|\cdot |K|}{|H\cap K|} = \frac{5\cdot 5}{1} = 25$, which is impossible in a group of $20$ elements. (I'm assuming you are familiar with that formula?) Therefore this situation is not possible and $|H\cap K| = 5$ indeed.

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    Is the formula you mentioned that gcd times lcm =a times b?And if this is the case, then this is pretty marvelous.2011-02-18
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    -1. The poster asked for hints and not for a complete solution. I don't think that doing somebody's homework for him is going to help him learn mathematics.2011-02-18
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    @Myself: Indeed, I took my nose away from the books after a night sleep. When I got back, I first read Arturo Magidin's comment and, surprisingly, I managed to write the same argument you just presented, albeit your's very much succinct and more easy to follow than mine. I appreciate your care for not telling things right and give spoiler alerts throughout your posts. Thanks!2011-02-18
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    @Alex Bartel: I appreciate your concerns (seriously, I do) but in this case there are spoiler alerts, so that's OK. Only after hours of frustration I put my arms down and take a look at some explanation. But thanks for your care.2011-02-18
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    @Alex: I think hints are great for in a classroom, where you can closely watch and steadily unveil more and more hints, until students manage to work out the problem by themselves. However, in written communication like email and this it makes more sense to alert the reader and encourage them to stop reading and try the rest themselves. Spoilertags would be very welcome for that purpose. But this is a bit of a metadiscussion, I guess.2011-02-18
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    @Myself (that address sounds kind of strange) A spoiler tag exists. You can search meta to find out how to use it, I don't have the link handy.2011-02-18
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    Please don't use that tag! Having things blink is the last thing we need! :/2011-02-18
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I know you stated that you have not learned the Sylow theorems, but here is an alternative method that you can look at whenever you're ready:

$ I. $ Let $p$ denote a prime and let $G$ be a group of order $n = p^am$ with $ a \in \mathbb{N}_0$ and $ m \in \mathbb{N}$ such that $p \not \mid m.$ Every subgroup of order $ p^a$ is called a Sylow $ p$-subgroup of $G.$ Here, we denote the set of Sylow $ p $-subgroups as $\operatorname{Syl}_p(G).$

$II. $ Let $G$ have order $n = p^am $ as described above. Then the number of Sylow $ p $-subgroups $|\operatorname{Syl}_p(G)|$ (this is all equal to the normalizer $N_p (G)) $ satisfies: $(i) |\operatorname{Syl}_p(G)| \equiv 1 \mod p $

$(ii) |\operatorname{Syl}_p(G)| \mid m$

(You can find plenty of proofs to $ II $ online. )

Now to your question:

We have $|G| = 20 = 5 \cdot 2^2.$ By $II. $ we have that $|\operatorname{Syl}_5(G)| \equiv 1 \mod 5$ and that $|\operatorname{Syl}_5(G)| \mid 2^2 = 4,$ but this implies that $|\operatorname{Syl}_5(G)| = 1.$ This means that there is only one such subgroup of order $5.$ Any group of order $20$ will contain only one subgroup of order $5,$ so any two subgroups $ K $ and $ H $ of $ G $ of order $5$ will be the same subgroup.

Edit: Didn't realize the date on the question, disregard any irrelevance