Let A be an abelian group, m a positive integer and $f_{m}:A{\rightarrow}A$ be defined by $f_{m}(a)=ma$. Show that $f_{m}$ is a group homomorphism.

Question

Let A be an abelian group and $f_{m}:A{\rightarrow}A$ be defined by $f_{m}(a)=ma$, with $m\in\mathbb{N}$, $m\geq1$.

• Show that $f_{m}$ is a group homomorphism and call $f_{m}(A)=mA$.
• Show $mA$ is a subgroup of $A$, and compute the kernel of $f_{m}$.
• Give an example that shows that if $A$ is not abelian, then $f_{2}$ is not an homomorphism.

I know we are dealing with symmetric groups, permutations, etc. but I'm really struggling to connect the dots. Also, I thought starting with the third point and finding an example would help me understand the two first points but I can't think of anything.

Answer 1

Hints.

1. We need to show that $$f(x+y)=f(x)+f(y)$$ for all $x,y\in A$. Can you finish from here?
2. Recall the definition of a subgroup.
3. Take any non-abelian group $G$ (for example, you mentioned that you know the symmetric group $S_3$ consisting of all permutations of the set $\{1,2,3\}$) and $a,b\in G$ such that $a+b\neq b+a$. Then $f_2(a+b)=a+(b+a)+b\neq a+(a+b)+b=f_2(a)+f_2(b)$.

For 3. concretely: Take the permutations (in cycle notation) $a=(1 \ 2 \ 3)$ and $b=(1 \ 3) (2)$ then $$b\circ a=(1\ 2) (3)\neq(1)(2\ 3)=a\circ b.$$