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Let $G$ be cyclic group, and $H_1, H_2$ subgroups. $|H_1|=15$, and $|H_2|=25$ Find $|H_1 \cap H_2|$.

So this is the solution we were presented at recital:

$|H_1|$ and $|H_2|$ divides $|G|$, so $|G|=lcm(15,25)=75k$, $k \in \mathbb N$. Since $G$ is cyclic, so are its subgroups, so $H_1=<\frac {|G|}{|H_1|}>=<\frac {75k}{15}>=<5k>$. Same with $H_2$ we get $H_2=<3k>$.

Then $H_1 \cap H_2=<lcm(5k,3k)>=<15k>$, and then we get $H_1 \cap H_2=<\frac{|G|}{|H_1 \cap H_2|}> \Rightarrow <15k>=<\frac{75k}{|H_1 \cap H_2|}> $, hence $|H_1 \cap H_2|=5$


But with a quick observation one can see that $|H_2 \cap H_2|$ divides both $|H_1|$ and $|H_2|$, so one can say $|H_2 \cap H_2|=gcd(|H_1|,|H_2|)=5$.

Is my solution correct? Is it correct always, or only when $G$ is cyclic?

And how could you then solve the question for non-cyclic $G$?

Thanks!

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3 Answers 3

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$\;G\;$ cyclic and finite (why?) , so it has one unique subgroup of each order dividing its order.

$$|H_1\cap H_2|\;\mid \;15\,,\,25\implies |H_1\cap H_2|=1,5$$

But there's a subgroup of $\;G\;$ of order $\;5\;$, and since any subgroup of a cyclic one is cyclic, this subgroup of order $\;5\;$ is a subgroup both of $\;H_1\;$ and of $\;H_2\;$, and we're done.

In case $\;G\;$ is not cyclic you cannot prove this (counterexample...?)

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  • $\begingroup$ Sorry, I didn't understand your reasoning to why it has to be of order 5. Why can't it be the trivial element only? Can you explain again? Thanks! $\endgroup$
    – Gilian
    May 6, 2014 at 16:30
  • $\begingroup$ "A subgroup of a cyclic group is cyclic". That $\;G\;$ has a subgroup of order five (and also one of order 3, say) is clear? Then this unique subgroup of order five of $\;G\;$ has to be contained in both $\;H_1\,,\,\,H_2\;$ ... $\endgroup$
    – DonAntonio
    May 6, 2014 at 16:34
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    $\begingroup$ Ah, yes. Very elegant solution! Thank you. $\endgroup$
    – Gilian
    May 6, 2014 at 16:36
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When $G$ is cyclic, of finite order $n$, the lattice of subgroups of $G$ is isomorphic to the lattice of the positive divisors of $n$, so a subgroup of order 15 and a subgroup of order 25 will intersect in a subgroup of order $\gcd(15, 25) = 5$.

Another answer has already shown that this does not necessarily work anymore if $G$ is not cyclic.

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  • $\begingroup$ Uhh, we haven't studies lattices yet. Can I prove that it's true in another way? Or it can be a way for me to check if I got the right answer? $\endgroup$
    – Gilian
    May 6, 2014 at 16:28
  • $\begingroup$ @Gilian, it simply means, first that given a positive divisor $k$ of $n$, then $G$ has a unique subgroup $K$ of order $k$. And then that given two divisors $h, k$ of $N$, and the corresponding subgroups $H, K$, then $h \mid k$ iff $H \le K$. $\endgroup$ May 6, 2014 at 16:33
  • $\begingroup$ @Gilian, you're welcome! $\endgroup$ May 6, 2014 at 16:39
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For non-cyclic $G$, you can't solve it unless you know something more about the subgroups than just their orders. For instance, if $G = \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$, then $G$ has two subgroups of order 2 with trivial intersection.

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  • $\begingroup$ Thank you, this can come in handy! $\endgroup$
    – Gilian
    May 6, 2014 at 16:40

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