Show that $G\oplus H$ is cyclic iff the finite groups $G$ and $H$ are cyclic and $\gcd(|G|,|H|)=1$
My answer is:
$(\Rightarrow )$ Suppose that $G\oplus H$ is cyclic. Then there is a generator $g=(g_1 ,g_2)$ of that group. For any elements $x\in G,y\in H $, $h=(x,y)\in G\oplus H$. So, $h=g^m$ for some positive integer $m$, i.e., $(g_1,g_2)^m=(g_1^m,g_2^m)=(x,y)$. Thus, $g_1,g_2$ generate G and H respectively.
Now, suppose for reductio that $\gcd(|G|,|H|)=a>1$. Let $|G|=a\cdot k$ and $|H|=a\cdot l$. Since $g_1,g_2$ are generator of G and H, $g^{a\cdot k\cdot l}=e_{G\oplus H}$. Then, since $akl<a^2kl$ by assumption $a>1$, g cannot generate $G\oplus H$. Contradiction.
($\Leftarrow$) Since each group is cyclic, there are generators $g_1,g_2$ of G,H respectively. So, if we take an element $g=(h_1,h_2)$, say $g_1^m=h_1, g_2^n=h_2$ of $G\oplus H$, then $(g_1,g_2)^{m\times n\times lcm(|G|,|H|)}=g$.
So, my questions are:
-
Is it correct?
-
In the right-left direction, it seems to me that the condition $\gcd(|G|,|H|)=1$ is useless. So, is it right to think that?
Best Answer
Your proof for $\Rightarrow$ is okay.
Your proof for $\Leftarrow$ is not okay.
E.g. test it on the abelian $\mathbb Z_2\oplus\mathbb Z_2$. Element $(1,1)$ has order $2$ hence does not generate it.
If $g\in G$ has order $n$ and $h\in H$ has order $m$ then the order of $(g,h)\in G\oplus H$ is the least common multiple of $n$ and $m$ which equalizes $n\times m$ iff $\gcd(m,n)=1$.
That is needed in order to prove that $G\oplus H$ is cyclic if $G$ and $H$ are.