# Let $H$ have order $m$ and $K$ have order $n$, where $m$ and $n$ are relatively prime. Then $H \cap K=\{e\}$

Let $H$ and $K$ be subgroups of $G$. Let $H$ have order $m$ and $K$ have order $n$, where $m$ and $n$ are relatively prime. Then $H \cap K=\{e\}$

My proof:

Let $H$ and $K$ be subgroups where the $ord(H)=m$ and $ord(K)=n$ and $m,n$ are relatively prime. We know that $H \cap K$ is a subgroup of $H$ and $K$ since it contains the elements of both in $H$ and $K$. We shall let the ord($H \cap K$)=d.

Then, by Lagrange's theorem, the order of $H$ is a multiple of the order of $H \cap K$. In other words, $m$ is a multiple of $d$ or $d|m$. Similarly, by Lagrange's theorem, the order of $K$ is a multiple of the order of $H \cap K$. In other words, $n$ is a multiple of $d$ or $d|n$. Since $d$ divides both $m$ and $n$ and we know $m$ and $n$ are relatively prime then, the order of $H \cap K$ must be $1$.

Since $H \cap K$ is a subgroup, then by properties of subgroup, it contains an inverse which means it must also contain an identity and since $H \cap K$ contains one element, it must contain the identity. As a result, $H \cap K=\{e\}$

Hopefully, someone can confirm or correct any mistakes I made. Thanks!

• It is correct, some of your expressions might be improved a bit though. For instance, at the very end, containing an identity isn't really an outcome of it being a subgroup rather than a necessity for it to be a subgroup. The point being you cannot define what an inverse is, if you don't have the identity in the first place... Apr 10, 2015 at 0:24
• That sounds better, thanks Theo! Apr 10, 2015 at 2:16

We have that $H \cap K \leqslant H \Rightarrow |H \cap K| \mid |H|$ (Lagrange) and similarly $H \cap K \leqslant K \Rightarrow |H \cap K| \mid |K|.$ Hence $|H \cap K|$ is a common divisor of both $|H|$ and $|K|.$ Since $\gcd (|H|,|K|) = 1\Rightarrow |H \cap K | = 1 \Rightarrow H \cap K = \{1\}. \text {} \Box$