$3\mathbb Z$ as a direct summand of $\mathbb Z$

Question

Show that $\mathbb Z$ cannot have $\mathbb Z$ as a direct summand: Suppose $G$ and $3\mathbb Z$ are the two direct summands of $\mathbb Z$ and show that $G$ must be isomorphic to $C_3$.

First of all, what exactly does the assumption say: $\mathbb Z=3\mathbb Z \oplus G$ or $\mathbb Z\simeq 3\mathbb Z \oplus G$?

Second, I have only an intuitive way of showing that $G\simeq C_3$. Namely, "mod out both sides by $\mathbb 3\mathbb Z\oplus\{0\}$". If I'm given that $\mathbb Z=3\mathbb Z \oplus G$, then the only problem I see is that formally speaking $\mathbb 3\mathbb Z\oplus\{0\}$ isn't a subgroup of $\mathbb Z$, so I need to say more words. What exactly the formal proof should look like? If I'm given $\mathbb Z\simeq 3\mathbb Z \oplus G$, then apart from the above think I should also justify that if two groups are isomorphic, then taking quotients by isomorphic (I guess) subgroups should yield the same result. Why is that true?

Answer 1

If $\mathbb Z \cong G\oplus 3\mathbb Z$, then there is canonical epimorphism $p\colon \mathbb Z\to G$ whose kernel is $3\mathbb Z$. Define map $\varphi\colon G\to\mathbb Z/3\mathbb Z$ by $\varphi (p(n)) = n + 3\mathbb Z$.

We need to show that this map is well defined, so let $p(m) = p(n)$. This implies that $p(m-n) = 0$, or $m-n\in\ker p = 3\mathbb Z$. Thus, $\varphi(p(m)) = m + 3\mathbb Z = n + 3\mathbb Z= \varphi (p(n))$.

$\varphi$ is obviously surjective, so it remains to show that it is injective. Also, I didn't show that it is homomorphism. Can you finish it?

Answer 2

What the assumptions says depends on whether you are talking about internal or external direct products. Since you are talking about $3\mathbb Z$ which already explicitly identified as a subgroup of $\mathbb Z$, it probably makes sense to talk about internal direct products here, in which case equality is appropriate.

These two notions are equivalent: Any time you write $G = H \oplus K$ as an internal direct product, this is equivalent to saying that $G \cong H \oplus K$ as an external direct product, via a particular isomorphism $\phi: G \to H\oplus K$ which satisfies $\phi(H) = H\oplus 0$ and $\phi(K) = 0\oplus K$.

If you consider external direct products instead, then just writing $\mathbb Z \cong 3\mathbb Z \oplus G$ doesn't fully describe the problem. It should be assumed there is not just any is isomorphism $\phi: \mathbb Z \to 3\mathbb Z \oplus G$, but one which additionally satisfies $\phi(3\mathbb Z) = 3\mathbb Z\oplus 0$.

As for "apart from the above ...", this is right, however you should be careful with what you mean by "isomorphic subgroups". If this just means "subgroups which are isomorphic to each other as groups" then the statement is false; for example $n\mathbb Z \cong \mathbb Z$ for all $n\neq 0$ in $\mathbb Z$, but $\mathbb Z / n\mathbb Z = C_n$ are not isomorphic for different values of $n$ (usually). Instead you need a choice of isomorphism between the subgroups which is compatible with the choice of isomorphism between the larger groups.

Here is a precise version:

"If $\phi: G \to H$ is an isomorphism of groups, and $G' \subseteq G, H' \subseteq H$ are (normal) subgroups such that $\phi(G') = H'$, then there is an induced isomorphism of groups $\phi': G/G' \to H/H'$ given in terms of coset by $\phi'(gG') = \phi(g)H'$."

As an aside: It would also be easy to show that $3\mathbb Z$ is not an internal direct summand of $\mathbb Z$ instead by showing that any subgroup $G$ of $\mathbb Z$ for which $G\cap 3\mathbb Z=0$ is trivial. (If $n \in G$, then $3n \in \mathbb Z\cap G = 0$ so $n=0$.)