If $H$ and $K$ are subgroups of prime order $p$, show that $H = K$ or $H \cap K = \{1\}$

Question

Say $G$ is a group and $p$ is a prime number. If $H$ and $K$ are subgroups of order $p$, show that $H = K$ or $H \cap K = \{1\}$

My thought would be the Lagrange Theorem. If $H$ is a subgroup of order $p$ of group $G$, then $|H|$ divides $|G|$. If $|G|$ is divided by two subgroups with the same order, then the result is the same. The condition where the subgroups are are the same makes sense (mostly), but the other condition makes no sense.

I honestly have no idea how one would prove this.

Edit: Prime group is cyclic. Cyclic group generated by a single element. If the groups aren't the same, $(H=K)$, then the only element they have in common is the identity element (in this case 1)?

Answer 1

Suppose $H\neq K$. We have that $H\cap K$ is a subgroup of both $H$ and $K$. By Lagrange, either $|H\cap K|=p$ or $|H\cap K|=1$, but if the former, then we must have $H=H\cap K=K$, a contradiction; hence $|H\cap K|=1$, which can only happen if $H\cap K=\{1\}$.

Note that $A\lor B$ is equivalent to $(\lnot A)\to B$.

Answer 2

Use the primality of $p$ and Lagrange's theorem to conclude that for all $h\in H $ such that $h\neq 1$ we have $$ H=\{h^{0}, h^{1},h^{2},\ldots, h^{p-1}\}. $$ In an entirely analogous way for all $k\in K $ such that $k \neq 1$ we have $$ K=\{k^{0}, k^{1},k^{2},\ldots, k^{p-1}\}. $$ Since $H\cap K$ is a group, if $u\in H\cap K$ and $u\neq 1$ then $u^{0}, u^{1},u^{2},\ldots, u^{p-1}\in H\cap K$ and $|H\cap K|=p$. Now,

• if $|K|=p$, $|H\cap K|=p$ and $H\cap K\subset K$ then $H\cap K=K$,
• if $|H|=p$, $|H\cap K|=p$ and $H\cap K\subset H$ then $H\cap K=H$.

Therefore, we can conclude that $H=K$.