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 $$\text{ord}(H)=m$$ and $$\text{ord}(K)=n$$ 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 $$\text{ord}(H \cap K)=d$$.

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 $\gcd (|H|,|K|) = 1\Rightarrow |H \cap K | = 1 \Rightarrow H \cap K = \{1\}. \text {} \Box$
• @Bluey Then $| H \cap K|$ is not necessarily the identity. Mar 7 at 17:20