How to prove $\langle x,y\rangle\cong\langle x\rangle+ \langle y\rangle$ in groups?
I am not sure if got the notations right, basically I was wondering given an additive group $G$, which is commutative, and two elements in $G$, $x$ and $y$, I was wondering if the subgroup generated by $x$ and $y$ would be isomorphic to the direct sum of $\langle x\rangle$ and $\langle y\rangle$.
I hope I have not messed up somewhere but I thought the natural map would be $\phi:
\langle x\rangle+ \langle y\rangle \to\langle x,y\rangle$ such that $\phi(u,v)=u+v.$ Now I can show this is group homomorphism and surjective quite easily, are there easy ways of showing this is also injective?
Many thanks!
Best Answer
It's not true. For instance, if you had $x=y$, this is clearly going to fail, since $\langle x,y\rangle$ would just be $\langle x\rangle$. However, you can get a related true statement out of this.
First, note that your definition of $\phi$ does not really make sense; the elements of $\langle x,y\rangle$ are elements of $G$, not pairs $(u,v)$. I think you might have your domain and codomain reversed - you might instead want the function $$\phi:\langle x\rangle \oplus \langle y\rangle \rightarrow \langle x,y\rangle$$ where, if we represent $\langle x\rangle \oplus \langle y\rangle $ to be the set of pairs $(u,v)$ with $u\in \langle x\rangle$ and $v\in\langle y\rangle$, we have $$\phi(u,v)=u+v.$$ I think this is probably what you meant, but we have to be precise. We can compute the kernel of this map. In particular, we get $$\phi(u,v)=0$$ if and only if $u+v=0$. The set of pairs $(u,v)\in \langle x\rangle \oplus \langle y\rangle$ looks like a copy of $\langle x\rangle \cap \langle y\rangle$, since if you pick any $u$ in this intersection, then $-u$ remains in the intersection (and if $u+v=0$ then $u=-v$ must be in $\langle y\rangle$ as well as $\langle x\rangle$).
If you use the map $\phi(u,v)=u-v$, which also works, you can very directly see that $$\langle x,y\rangle = \frac{\langle x\rangle \oplus \langle y\rangle}{\Delta(\langle x\rangle \cap \langle y\rangle)}$$ where $\Delta$ is the embedding of $\langle x\rangle \cap \langle y\rangle$ into $\langle x\rangle \oplus \langle y\rangle$ taking $g$ to $(g,g)$ - the point being that the join of the groups is the direct sum mod the intersection. This is, fairly often, a nice fact to know - it's similar, though not quite identical, to the second isomorphism theorem.