Is order of Image of $A\times B$ under group homomorphism is smaller than that of $B$ when there is no nontrivial homomorphism from $A$ to target?

Question

I will explain the title of this question from now.

Let $A,B,C$ be abelian groups. Suppose $B,C$ are finite abelian groups.

Suppose there does not exists a nontrivial group homomorpshim from $A$ to $C$.

Let $f: A\times B \to C$ be a any fixed group homomorposhim.

Is it true that　order of $f(A\times B)$(order of image of $A\times B$ under $f$) is smaller than order of $ B$ ?

What in my mind is the case ,for example, $A=\Bbb{Q}$, $B=\Bbb{Z}/16\Bbb{Z}$,and$C$ is a appropriate finite group.

Answer 1

The direct product of two groups (not necessarily abelian) has the following universal property:

Theorem. Let $G$ and $H$ be groups. Let $i_G\colon G\to G\times H$ be the morphism $i_G(g) = (g,e)$, and let $i_H\colon H\to G\times H$ be the morphism $i_H(h) = (e,h)$. Then for every group $K$ and any morphisms $f_1\colon G\to K$, $f_2\colon H\to K$, if $f_1(g)$ commutes with $f_2(h)$ for every $g\in G$ and $h\in H$, then there exists a unique morphism $\Phi\colon G\times H\to K$ such that $f_1 = \Phi\circ i_G$ and $f_2=\Phi\circ i_H$. Conversely, any morphism $\Psi\colon G\times H\to K$ corresponds to a pair of morphisms $\psi_G\colon G\to K$ and $\psi_H\colon H\to K$ such that $\psi_G(g)$ commutes with $\psi_H(h)$ for each $g\in G$ and $h\in H$, and $\Psi(g,h) = \psi_G(g)\psi_H(h)$ for all $(g,h)\in G\times H$. Moreover, the two correspondences are inverses of each other.

Proof. Given $f_1$ and $f_2$, define $\Phi$ by $\Phi(g,h) = f_1(g)f_2(h)$. This is a morphism because the images of $G$ and $H$ commute in $K$: $$\begin{align*} \Phi\Bigl( (g_1,h_1)(g_2,h_2)\Bigr) &= \Phi\Bigl( g_1g_2,h_1h_2\Bigr)\\ &= f_1(g_1g_2)f_2(h_1h_2)\\ &= f_1(g_1) f_1(g_2) f_2(h_1) f_2(h_2)\\ &= f_1(g_1) f_2(h_1) f_1(g_2) f_2(h_2)\\ & \Phi(g_1,h_1) \Phi(g_2,h_2). \end{align*}$$ Moreover, $\Phi\circ i_G(g) = \Phi(g,e) = f_1(g)f_2(e) = f_1(g)$, so $\Phi\circ i_G = f_1$, and symmetrically $\Phi\circ i_H(h)=\Phi(e,h) = f_1(e)f_2(h) = f_2(h)$, so $\Phi\circ i_H=f_2$. Uniqueness follows because $G\times H$ is generated by $i_G(G)$ and $i_H(H)$.

Conversely, given $\Psi\colon G\times H\to K$, let $\psi_G\colon G\to K$ be defined by $\psi_G(g) = \Phi(g,e)$ (that is, $\psi_G=\Psi\circ i_G$) and $\psi_H = \Psi\circ i_H$. They are group homomorphisms since they are the result of composing group homomorphisms. We also have for all $g\in G$ and $h\in H$ that $$\begin{align*} \psi_G(g)\psi_H(h) &= \Psi(g,e)\Psi(e,h)\\ &= \Psi\Bigl( (g,e)(e,h)\Bigr)\\ &= \Psi(g,h)\\ &= \Psi\Bigl( (e,h)(g,e)\Bigr)\\ &= \Psi(e,h)\Psi(g,e)\\ &= \psi_H(h)\psi_G(g), \end{align*}$$ So $\psi_G(g)$ commutes with $\psi_H(h)$. And the same calculation as above shows that $\Psi(g,h) = \Psi(g,e)\Psi(e,h) = \psi_G(g)\psi_H(h)$, as desired.

It is now straightforward to verify the two corerspondences are inverses of each other.$\Box$

In particular:

Corollary. If $G$ and $H$ are groups, and $C$ is an abelian group, then morphism $f\colon G\times H\to C$ correspond to pairs of morphisms $g\colon G\to C$ and $h\colon H\to C$; the corespondence associates the morphism $f$ to the unique pair $(g,h)$ such that $f(x,y) = g(x)h(y)$ for all $(x,y)\in G\times H$.

From this we get that if the only morphism $A\to C$ is trivial, then for every morphism $f\colon A\times B\to C$ we have $\mathrm{Im}(f) = f(\{e\}\times B) = h(B)$, where $h$ is the corresponding map from the corollary. So in your situation, the image $f(A\times B)$ is always the same as the image $f(\{e\}\times B)$, and therefore is isomorphic to a quotient of $B$. So the size is always at most the size of $B$, but could equal the size of $B$.