# How to prove $\langle x,y\rangle\cong\langle x\rangle+ \langle y\rangle$ in groups?

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!

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.
The element $$(u,v)$$ is not an element of $$\langle x,y\rangle$$... Instead, an arbitrary element of $$\langle x,y \rangle$$ is of the form $$ax + by$$, where $$a, b \in \mathbb Z$$. Then you can define $$\phi$$ by $$\phi(ax + by) = (ax, by)$$. It should be pretty easy to show that this is an isomorphism.