# Proof question about Find the necessary and sufficient condition such that $\langle (g,h) \rangle = \langle g \rangle \oplus \langle h \rangle$

Let $G$ and $H$ be finite groups and let $(g,h) \in G \oplus H$. Find the necessary and sufficient condition such that $\langle (g,h) \rangle = \langle g \rangle \oplus \langle h \rangle$ where $\oplus$ is the external direct product.

I've figured that gcd$(|g|, |h|) = 1$ should probably be the condition, but I'm stuck at a specific part of the proof.

I've shown that $\langle (g,h) \rangle \subseteq \langle g \rangle \oplus \langle h \rangle$ is true without gcd$(|a|, |b|) = 1$ and now I'm trying to show that $\langle g \rangle \oplus \langle h \rangle \subseteq \langle (g,h) \rangle$.

I know this means I have to prove that $(\forall(x,y) \text{ in } \langle (g,h) \rangle)(\exists \in \Bbb Z)(g^b=x \land h^b = y)$.

Letting $(x,y)$ be an arbitrary element in $\langle g \rangle \oplus \langle h \rangle$, this means that there are $n,m \in \Bbb Z$ such that $g^n = x \land h^m = y$. I figure in order to show this I have to multiply $g^n$ and $h^m$ by enough factors such that $a = b$ for $g^a = x$ and $h^b = y$, but I keep getting stuck here.

• What kind of things are $g$ and $h$, and what does the $\langle\cdots\rangle$ mean? – Henning Makholm Aug 3 '16 at 15:38
• Edited for clarity. – Oliver G Aug 3 '16 at 15:40
• I think you have to prove that a $b$ exists for all $x\in\left<g\right>, y\in\left<h\right>$, not just for $(x,y)\in\left<(g,h)\right>$. (The later is immediate by definition of $\left<\cdots\right>$ and the elementwise operation in the product group). – Henning Makholm Aug 3 '16 at 15:45
• And then what you need is the Chinese Remainder Theorem. – Henning Makholm Aug 3 '16 at 15:46

## 1 Answer

Let $|g|=n,|h|=m$.

Notice $\langle g,h\rangle =\{(g^n,h^n)\mid n\in \mathbb Z\}$. This set has $|(g,h)|=\mathrm{lcm}(n,m)$ elements.

On the other hand $\langle g \rangle \oplus \langle h \rangle =\{(g^n,h^m)\mid n,m\in\mathbb Z\}$. This set has $nm$ elements and clearly contains the previous set.

We conclude that the sets are equal if and only if they have the same size, that is, $nm=\mathrm{lcm}(n,m)=\frac{nm}{\gcd(n,m)}\iff n$ and $m$ are coprime.