let f be a homomorphism from a finite group G into the Q/Z then prove that

Question

let f be a homomorphism from a finite group G into Q/Z, where (Q,+) is additive group of rationals. Then prove that

(a) For all g∈G,f(g) is of finite order.

(b) f(G) is a cyclic subgroup of Q/Z.

(c) If f is one-one then every subgroup of G is normal.

(d) if o(f(G)) = n then G must have an element of order n.

Source: My old test papers of group theory

my try for (a) since Q/Z has all proper subgroup of finite order then may be (a) is proved. But stuck on (b) becouse f(G) can be union of cyclic subgroup. for (c) no idea and for (d) o(f(a)) divides o(a).

Answer 1

(a) $f(g)$ is of the form $\overline{\frac{p}{q}}$, hence it is of finite order (in particular $f(g)^q=1$).

(b) Let $G=\{g_1,\dots,g_k\}$ and assume $f(g_i)=\overline{\frac{p_i}{q_i}}$ for all $i$. Then $f(G)$ is contained in the cyclic subgroup generated by $\overline{\frac{1}{q}}$ with $q=\prod_{i=1}^k q_i$. Hence $f(G)$ is cyclic.

(c) If $f$ is injective, then $G$ is a subgroup of an abelian group, hence is abelian and so every subgroup is normal.

(d) What do you mean by $o(f(G))$?

Answer 2

(a) For all $g\in G$,$f(g)$ is of finite order.

That is because $f(g)$ belongs to $\Bbb{Q}/\Bbb{Z}$ where all elements have finite orders: $(a/b +\mathbb{Z})b=\Bbb Z$.

(b) $f(G)$ is a cyclic subgroup of $\Bbb{Q}/\Bbb Z$.

This is because every finite subgroup of $\Bbb{Q}/\Bbb Z$ is cyclic (which follows from the fact that every finitely generated subgroup of $\Bbb{Q}$ is cyclic).

(c) If $f$ is one-one then every subgroup of $G$ is normal

because then $G$ is isomorphic to $f(G)$ and every subgroup of a cyclic group is normal.

(d) if $o(f(G)) = n$ then $G$ must have an element of order $n$.

Since $f(G)$ is cyclic it is generated by one element $a$. Let $g\in G$ such that $f(g)=a$. Then $|f(\langle g\rangle)|=n$. By Lagrange then $n$ divides $o(g)$. Then $\langle g\rangle$ contains a subgroup of order $n$. Since every subgroup of a cyclic group is cyclic, you get the result.