Let $A,B,$ and $C$ be multiplicative groups, let $f:A→B$ and $g:A→C$ be group homomorphisms such that $\ker f ∩ \ker g=\{1\}.$ Let $F:A→B×C$ be defined by $F(a)=(f(a),g(a))$.
(a) Show that $\ker F$ is trivial.
(b) Find a subgroup of $B × C$ that is isomorphic to $A$.
My attempt:
First, we can show that $F$ is also a group homomorphism:
$F(a_1a_2)=(f(a_1a_2),g(a_1a_2))=(f(a_1)f(a_2),g(a_1)g(a_2))$ (since $f:A→B$ and $g:A→C$ are group homomorphisms) $=(f(a_1),g(a_1))(f(a_2),g(a_2))$ (the operation on the direct product $B×C$ is component-wise) $=F(a_1)F(a_2)$.
Now, for part a) I tried to use the fact that the kernel of a group homomorphism $F$ is trivial if and only if $F$ is injective. So suppose $F(a_1)=F(a_2)$, then $$(f(a_1),g(a_1))=(f(a_2),g(a_2)) \iff \begin{cases} f(a_1)=f(a_2) \\ g(a_1)=g(a_2) \end{cases}.$$ But I don't think this is the correct way to prove it, I believe we need to use the fact that $\ker f ∩ \ker g=\{1\}$ somehow. $\ker f ∩ \ker g=\{1\}$ doesn't imply $\ker f=\ker g=\{1\}$ right?
As for part b) I am completely stuck and have no ideas whatsoever.
Any help would be appreciated.
Best Answer
Let $x$ be in $\textrm{Ker}(F)$. As the neutral element of $B \times C$ is $(1_B, 1_C)$, one has $(f(x), g(x)) = (1_B, 1_C)$. So,$x$ belongs to both $\textrm{Ker}(f)$ and $\textrm{Ker}(g)$. Thus, $x = 1_A$ and $\textrm{Ker}(F)$ is trivial because included in $\{ 1_A \}$ (the reciprocal inclusion is obvious).
This fact is a characterisation of the injectivity of group homomorphisms. This is sufficient to claim that the kernel of $F$ is included in $\{ 1_A \}$.
As $F$ is injective, $F(A)$ is isomorphic to $A$ and as $F$ is a group homomorphism, $F(A)$ is a subgroup of $B \times C$. So $F(A)$ is the subgroup you are looking for.