2
$\begingroup$

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!

$\endgroup$

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .