How to prove set inclusion in a group with a prime cardinality?

Question

Given the question:

If $H,\,K$ are subgroups of a group $G$, and $\lvert H\rvert$ is a prime number, show that either $H\subseteq K$ or $H\cap K=\{1\}$.

I tried solving it as follows:

Assume $K\subset H$ then $K$ is a group and a subset of $H$, so $K$ is a subgroup of $K$.

Then by Lagrange's theorem, $|K|\mid |H|$, yet $|H|$ is a prime number so $|K|=1$. Since $K$ is a group it must contain the identity so $K=\{1\}$.

Assume $K \not \subset H$, then either $H \subseteq K$ or $\exists k \in K$ and $\exists h \in H$ such that $k \notin H$ and $h \notin K$.

Assume the second, then we have that, since $H \subseteq G$ and $K \subseteq G$ then $kh = z$ for some $z \in G$.

If $z \in H$ we have $k = zh^{-1}$ and since $h,h^{-1} \in H$ and $H$ is a group, then $k \in H$.

Which is a contradiction.

For the same argument $z \notin K$.

However all this lets me conclude is that $z \notin K$ and $z \notin H$

Which isn't particularily useful. I wonder if there is a hint as to what I could try to finalize the proof. Maybe a property of groups I skipped.

Answer 1

Suppose

$H \cap K \ne \{ e \}, \tag 1$

where $e \in G$ is the identity element. Then

$\exists \; e \ne h \in H \cap K; \tag 2$

since $\vert H \vert = p$ is prime, $H$ is a cyclic group of order $p$ and hence is generated by any non-identity element; thus we may write

$H = \langle h \rangle; \tag 3$

now since $h \in K$, and $K$ being a subgroup is closed under the group operation, every power $h^n$ of $h$ is in $K$; thus

$H = \langle h \rangle \subset K; \tag 4$

Note Added in Edit, 27 February 2018 11:23 PM PST:

Let $\vert H \vert = p$, a prime. Let $e \ne h \in H$ and consider the sequence of powers of $h$:

$h^1 = h,\; h^2,\; h^3, \; \ldots, h^p; \tag 5$

if all the entries in this list are distinct, then since $\vert H \vert = p$, every element of $H$ occurs precisely once; thus we must have $h^p = e$; if not then $h^m = e$ for some $m$, $1 \le m < p$; but then $h = he = hh^m = h^{m + 1}$ and so $h$ occurs at least twice in the list; hence $h^p = e$; since the list elements are assumed distinct, we have $H = \langle h \rangle$, the cyclic group generated by $h$. If the list entries are not distinct, then we must have $h^j = h^k$ for some $1 \le j < k \le p$; then $h^{k - j} = e$ and the elements

$h = h^1, \; h^2, \ldots, h^{k - j} = e \tag 6$

form a subgroup $H'$ of $H$; then by LaGrange's theorem, $\vert H' \vert \mid \vert H \vert$; since $\vert H \vert = p$ is prime, $\vert H' \vert = 1$ or $\vert H' \vert = p$; since $h \ne e$ we cannot have $\vert H' \vert = 1$, so $\vert H' \vert = p$; but this is impossible since the list (6) contains at most $p - 1$ entries. Therefore this assumption that the elements of the list (5) are not distinct may be ruled out.

If $x \in H$ with $x^2 = e$, then $\{1, x \}$ forms a subgroup of $H$ order $2$; again by LaGrange's theorem, $2 \mid p$. Since $p$ is prime, we must thus have $p = 2$ and $H = \langle x \rangle = \{ 1, x \}$.

End of Note.

Answer 2

Take any non-identity element $h \in H$ and let $|H|=p$. The order of $h$ divides $p$ (why?), and because $p$ is prime and $|h|>1$ we have $|h|=p$; thus $H= \left< h \right>$ (why?). Either $H\cap K = \{1 \}$ or $H\cap K \not = \{1 \}$. If the first is true, we are done. If the second is true, then $K$ contains a non-identity element $h \in H$ which will generate all of $H$ inside $K$, so $H \subseteq K$.

Answer 3

I would take an even easier approach. Take $H\cap K$. Since $H$ and $K$ are both subgroups $H\cap K$ is a subgroup of $H$. Then $|H\cap K|$ divides $|H|$ and so is either $1$ or $p$. If it's $1$ it must be $e$ if it's $p$ then all of $H=H\cap K\subseteq K$. We are using the fact that $p$ is finite in the second part.