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

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)?

• The interesction of two subgroups $H$ and $K$ is a subgroup of both $H$ and of $K$. Commented Nov 11, 2021 at 23:24
• $1$ is the identity element. Cyclic groups don't have to be "generated by numbers": groups are abstract objects, and their elements don't have to be numbers. Cyclic groups are generated by a single element. What "$H\cap K=\{1\}$" means is that the intersection consists only of the identity element. Commented Nov 11, 2021 at 23:26
• So a group of prime order would have two distinct subgroups: the trivial subgroup and the whole group. If H = K is false, then the subgroup is the trivial subgroup? Commented Nov 11, 2021 at 23:26

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 $$(\lnot A)\to B$$.
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 $$H=K$$.