Let $G$ be any abelian group and $a\in{G}$. Show there exists a homomorphism $f:G\rightarrow{\mathbb{Q}/\mathbb{Z}}$ such that $f(a)\neq{0}$. Let $G$ be any abelian group and $a\in{G}$. Show there exists a homomorphism $f:G\rightarrow{\mathbb{Q}/\mathbb{Z}}$ such that $f(a)\neq{0}$.
I can prove this question (I think) if I use the fact that $\mathbb{Q}/\mathbb{Z}$ is an injective abelian group: just define $f$ on the cyclic subgroup generated by $a$ and extend to $G$ via injectivity. However, I feel there should be a more 'elementary' way to prove this result, I just can't see one yet. One of my friends suggested using Zorn's Lemma, but I haven't done much with this information yet.
 A: Let $M$ be an abelian group.
Every integer $n$ defines an endomorphism $f_n\colon M \rightarrow M$ such that $f_n(x) = nx$.
$M$ is called divisible if $f_n$ is surjective for every nonzero integer $n$.
Clearly  $\mathbb{Q}/\mathbb{Z}$ is divisible.
Lemma 1
Let $M$ be a divisible abelian group.
Let $I$ be an ideal of $\mathbb{Z}$.
Then the canonical homomorphism $Hom(\mathbb{Z}, M) \rightarrow Hom(I, M)$ induced by the canonical injection $I \rightarrow \mathbb{Z}$ is surjective.
Proof:
If $I = 0$, the assertion is clear.
Hence we assume $I \neq 0$.
There exists a nonzero integer $n$ such that $I = \mathbb{Z}n$.
Let $f \in Hom(I, M)$.
Since $M$ is divisible, there exists $a \in M$ such that $f(n) = na$.
Let $x \in I$.
There exists an integer $m$ such that $x = mn$.
$f(x) = f(mn) = mf(n) = mna = xa$.
Hence $f$ is in the image of the map$\colon Hom(\mathbb{Z}, M) \rightarrow Hom(I, M)$.
QED
Lemma 2
Let $T$ be a divisible abelian group.
Let $M$ be an abelian group.
Let $N$ be a subgroup of $M$.
Let $f\colon N \rightarrow T$ be a homomorphism.
Let $x \in M - N$.
Then there exists a homomorphim $g\colon N + \mathbb{Z}x \rightarrow T$ extending $f$.
Proof:
Let $I = \{a \in \mathbb{Z}\colon ax \in N\}$.
Let $h\colon I \rightarrow T$ be the map defined by $h(a) = f(ax)$.
Since $h$ is a homomorphism, by Lemma 1, there exists $z \in T$ such that $h(a) = az$ for all $a \in I$.
Suppose $y + ax = y' + bx$, where $y, y' \in N, a, b \in \mathbb{Z}$.
$y - y' = (b - a)x$
Hence $b - a \in I$.
Hence $h(b - a) = f((b - a)x) = (b - a)z$.
Hence $f(y - y') = (b - a)z$.
Hence $f(y) + az = f(y') + bz$.
Therefore we can define a map $g\colon N + \mathbb{Z}x \rightarrow T$ by $g(y + ax) = f(y) + az$.
Clearly $g$ is a homomorophism extending $f$.
QED
Theorem
Let $T$ be a divisible abelian group.
Then $T$ is injective.
Proof:
This follows immediately from Lemma 2 and Zorn's lemma.
Corollary
$\mathbb{Q}/\mathbb{Z}$ is injective.
A: Here is a proof using Zorn's lemma: Define a non-trivial map $f_0: \langle a \rangle \to \mathbb Q/\mathbb Z$ on the cyclic subgroup generated by $a$, and consider pairs $(U,f)$ where $U$ is a subgroup of $G$ containing $a$ and $f: U \to \mathbb Q/\mathbb Z$ is a homomorphism extending $f_0$. Those pairs form a partially ordered set with the relation
$$(U,f) \leq (V,g) \Leftrightarrow U \subseteq V \text{ and } g\vert_U = f.$$
Clearly, every chain of such pairs has an upper bound. So, by Zorn's lemma, there is a maximal element $(U,f)$. We will show $U = G$. Suppose for contradiction that there is a $x \in G\setminus U$. We are finished if we can extend $f: U \to \mathbb Q/\mathbb Z$ to a homomorphism $\hat f: U + \langle x \rangle \to \mathbb Q/\mathbb Z$ since that would contradict the maximality of $(U,f)$. If $U \cap \langle x \rangle = 0$, the sum $U + \langle x \rangle$ is direct and we can just define $\hat f$ to be trivial on $x$. Otherwise, we have $U \cap \langle x \rangle = \langle nx \rangle$ for some integer $n \neq 0$ dividing the order of $x$. Our map $\hat f$ has to satisfy
$$n \hat f(x) = \hat f(nx) = f(nx),$$
so we take any $\alpha \in \mathbb Q/\mathbb Z$ satisfying $n \alpha = f(nx)$ (which exists as $\mathbb Q/\mathbb Z$ is divisible) and define $\hat f(x) := \alpha$. Using that $n$ divides the order of $x$, check that this extends to a unique homomorphism $\hat f: \langle x \rangle \to \mathbb Q/\mathbb Z$. It coincides with $f: U \to \mathbb Q/\mathbb Z$ on the intersection $U \cap \langle x \rangle = \langle nx \rangle$, so the two can be put together to a homomorphism $\hat f: U + \langle x \rangle \to \mathbb Q/\mathbb Z$. This is then a well-defined homomorphism extending $f$, giving the desired contradiction.
