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This question is from the counting principle section of my abstract algebra textbook. I understand the proof that is provided in my textbook. The proof uses a counterexample with subgroups $H$ and $K$ to get a contradiction with $o(H\cap K)=o(H)=o(K)$. I was wondering if this was an alternative proof.

Suppose there are two subgroups with order $p$. We know that $$o(H\cap K)\ge\frac{o(H)o(K)}{o(G)}>1$$

The middle equation would be $\frac{p^2}{pq}=\frac{p}{q}$. Since $p$ is a prime and $p>q$, $\frac{p}{q}$ wouldn't be an integer.

If I am wrong, I think I am incorrectly assuming that $\frac{p}{q}$ has to be an integer, but it feels like it has to, intuitively.

Musical Taco
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    $o(H \cap K) \geq p/q > 1$. That is correct. Why does it matter whether or not $p/q$ is an integer? It doesn't matter. Observe that $o(H \cap K)$ divides $p$, so... – azif00 Mar 25 '24 at 07:23
  • I guess I had index($H$)=$\frac{o(G)}{o(H)}$ in my head when I thought dividing orders led to integers. Is this proof a dead end? or is there something I can do that doesn’t match the standard proof? – Musical Taco Mar 25 '24 at 08:54
  • I am only aware of a proof of this using Sylow's theorem , but I guess you want a solution without this. – Peter Mar 25 '24 at 09:01
  • @Peter Sylow’s is coming up in a few weeks. My class just finished homomorphisms. – Musical Taco Mar 25 '24 at 09:02
  • Observe that in $D_{2p}$ for prime $p>2$, $p/2$ is not an integer. – Shaun Mar 25 '24 at 10:59
  • @Shaun Can you explain what $D_{2n}$ is? I know that you are just setting $n=p$, but I am confused about the difference between $D_n$ and $D_2n$. (Does the 2 just represent the order of a reflection dihedral subgroup?) – Musical Taco Mar 25 '24 at 17:16
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    Here I mean the dihedral group of order $2p$. Some people denote it $D_p$. Unfortunately, there is no agreed upon preference between the two, @MusicalTaco. – Shaun Mar 25 '24 at 17:18

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Let $H$, $K$ be subgroups of order $p$. Observe that the intersection $H\cap K$ of two subgroups is again a subgroup, and therefore its order divides $p$. But $p$ is prime, so we are left with two possibilities: either $o(H\cap K) = p$, in which case $H = K$ and we are done. The other case is $o(H\cap K) = 1$, which means the intersection of $H$ and $K$ is trivial. But then we have $o(HK) = p² > pq = o(G)$, a contradiction. (Note that $HK$ is seen only as a subset of $G$, not a subgroup).

Your proof leads to the same conclusion: If $o(H\cap K) > 1$, then it must be equal to $p$, and therefore $H = K$. It does not matter if the fraction evaluates to an integer, but I would like to know where the formula comes from.

paulina
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    Their proof doesn't hold for, say, dihedral groups. – Shaun Mar 25 '24 at 11:03
  • I got the formula from my lecture notes, but we have to write pretty fast. After class, I have to try and make out what my original notes were trying to say. If this formula makes no sense, I very well could have wrote something down incorrect. – Musical Taco Mar 28 '24 at 22:52