# Prove that $G$ has a subgroup $L$ and $|L|=mn$

For $G$ is an abelian group, $H,K$ are subgroups of $G$ and $|H|=n,|K|=m$. Prove that $G$ has a subgroup $L$ and $|L|=mn$

In cases $H\cap K=\{e\}$ use Lagrange theorem we can show that $|HK|=mn$ but in general i can't finish, can i help me?

Source : I.N. Herstein "Topics in Algebra", in this book Herstein ask: "Prove that a subgroup has order $lcm(m,n)$ with $H,K$ are subgroups of $G$ abelian group, and $|H|=n,|K|=m$.

• you need to qualify the assumption in some way. your first sentence would allow $H=K=G$ which would result in the assertion that $|G|=|G|^2$ – David Holden Feb 2 '15 at 17:29
• You need more hypothesis, what prevents $H=K=G$ or similar such things, I think $n$ and $m$ must be relatively prime. Then the proof is as you have given it. – Rene Schipperus Feb 2 '15 at 17:30
• Are you sure the problem does not ask for a group whose order is $lcm (m,n)$? – Ishfaaq Feb 2 '15 at 17:30
• The claim, as it is now, is painfully false: take for example the abelian group $\;C_2\times C_4\;$ , which has two subgroups $\;H,K\;$ of order $\;4\;$, yet it obviously has no subgroup of order $\;4\cdot4=16\;$ . – Timbuc Feb 2 '15 at 17:35

As has been shown here if we define a cyclic group $\{a, a^2, ... , a^{12}\}$ then taking $H = (a)$ and $K = (a^2)$ would mean there is no subgroup such that this problem asks to find.
However in one of the early sections in Herstein there is a similar problem that asks to show the existence of a subgroup of order $lcm (m,n)$ given two subgroups $H, K$ in $G$, an abelian group, such that $o(H) = m$ and $o(G) = n$. The desired group is $HK = \{hk \ | \ h \in H, k \in K\} = KH$.
$o(HK)$ exists in $\Bbb N$ since for any element $hk$ in it $(hk)^{mn} =(h)^{mn}(k)^{mn} = e$. Take $o(HK) = d$ and we can show that $d$ is the least common multiple of $m,n$ or that it is equal to $\frac{mn}{\gcd(m,n)}$.
It just remains to be seen that $(hk)^d = (h^{m})^{\frac{n}{\gcd(m,n)} } (k^{n})^{\frac{m}{\gcd(m,n)}} = e$ and that $(hk)^N = e$ means $h^Nk^N = e$ and so without loss of generality $h^N = k^N = e \implies m, n \ | \ N \implies d \ | \ N$.