# 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.

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.

• The only possible orders would be $1$ and $13$.
– lulu
May 25 at 20:37
• That goes not just for the order of the elements in the subgroup $H \cap K$, but for the order of the subgroup itself. May 25 at 20:40

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.
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, ...
• 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 $<x>$, and since $H$ contains $<x>$ the intersection is $<x>$. May 25 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? May 26 at 16:22
$$H\cap K$$ 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.