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Question: Suppose $G$ is a group and $H$ and $K$ are subgroups of $G$ such that $|H|=39$ and $|K|=65$. Prove that $H \cap K$ is cyclic.
[It is Question-$9$ of Chapter $10$ of Abstract Algebra by Dan Saracino]

Up to this chapter, Sylow Theorem has not been introduced, and we should solve this by Lagrange Theorem or Conjugacy Classes.

I solved this, but I think my solution is too complex. I write my answer here. Would you tell me if there is a better solution for this question and whether my solution is correct or not?

My solution:

Possible order for elements of $H$ are : $1, 3, 13, 39$

Possible order for elements of $K$ are : $1, 5, 13, 65$

If $H$ or $G$ do not contain an element of order $13$ then the intersection would be $\{e \}$ which is cyclic. So, I investigate the case where there is an element of order $13$ in both subgroups.

Now I am going to prove $H$ can not contain two or three cyclic subgroups of order $13$. If it contains an element of order $13$, it must only contain one subgroup of order $13$.

In the following different cases, $o(x_i)=13$ and $o(y_i)=3$:

Case 1: $H=\{e,x_1,x_1^2,\cdots,x_1^{12},x_2,x_2^2,\cdots,x_2^{12},x_3,x_3^2,\cdots,x_3^{12},y,y^2 \}$

Case 2: $H=\{e,x_1,x_1^2,\cdots,x_1^{12},x_2,x_2^2,\cdots,x_2^{12},y_1,y_2^2,\cdots,y_7,y_7^2 \}$

Case 3: $H=\{e,x_1,x_1^2,\cdots,x_1^{12},y_1,y_1^2,\cdots,y_{13},y_{13}^2 \}$

Case 1 and 2 are not possible because:

$\langle x_1,x_2\rangle \subset H$ and $|\langle x_1,x_2\rangle|>39$ which is contradiction.

Hence, only Case 3 is possible, and, indeed, in this case, $H=\{e,x,x^2,\cdots,x^{12},y,y^2,xy,xy^2,x^2y,x^2y^2,\cdots,x^{12}y,x^{12}y^2 \}$

Now, if $K$ does not contain $x$, then $H \cap K=\{ e\}$, which is cyclic. If $K$ contains $x$ then $H \cap K=\{ e,x,x^2,\cdots,x^{12} \}$, which is cyclic.

Rohit Singh
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khashayar
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3 Answers3

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Since $H,K\le G$, we have $H\cap K\le G$; therefore, in particular, $H\cap K$ is a group. By Lagrange's Theorem, the possible orders of $H\cap K$ are $1$ and $13$. The case $1$ is trivial. If the order is $13$, then, since $13$ is prime, $H\cap K$ is cyclic.

Shaun
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I do not quite follow your proposed proof (how do you come to those cases? why if $K$ contains $x$ then ...?), but as pointed out in comments, this exercise has a much easier proof:

The order of the group $H\cap K$ must divide both orders of $H$ and $K$. So it can only be ... or ..., and in either case, ...

Torsten Schoeneberg
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  • Let's say $C_i$ are cyclic subgroups and $\bar{C_i}=C_i - {e }$. I came up with those cases because I consider a group should be a union of disjoint $\bar{C_i}$ and ${e }$. Also if $K$ contains $x$, then it should contain $$, and since $H$ contains $$ the intersection is $$. – khashayar May 25 '22 at 21:15
  • Even if you mean proper cyclic subgroups, your first assertion does not hold already for $\mathbb Z/4$ or $\mathbb Z/6$. For the second argument, well why is the intersection not bigger? – Torsten Schoeneberg May 26 '22 at 16:22
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$H\cap K$ being a subgroup of $H$, its order must divide $39$, and being a subgroup of $K$, its order must divide $65$ as well. The only two numbers that divide both $39$ and $65$ are $1$ and $13$. The trivial group is cyclic, and any group of prime order is cyclic, too.

uniquesolution
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