# 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.

• What do you mean by "elementary"? – Makoto Kato Oct 15 '12 at 20:44
• maybe that was the wrong word to use, basically it's from an exercise in a book that asks this question as part (a) and then asks to prove $\mathbb{Q}/\mathbb{Z}$ is injective later in the question, so I really mean without the use of any homological algebra. A proof using group theory or set theory (like the Zorn's lemma suggestion) is what I meant when I said elementary. – Paul Gilmartin Oct 15 '12 at 20:47
• Of course, the group $\,G\,$ must be non-trivial and also $\,a\neq 0\,$ – DonAntonio Oct 15 '12 at 21:40

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.

• Very nice, thanks! – Paul Gilmartin Oct 15 '12 at 22:32
• It seems to be true (I think) that this proof only really needs the divisibility of $\mathbb{Q}/\mathbb{Z}$, which is equivalent to injectivity for abelian groups, so should hold for all injective abelian groups. – Paul Gilmartin Oct 15 '12 at 22:38

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.