Question about homomorphisms I have a question that asks the following:

Let $S,*$ and $T,.$ be binary structures and let the there be a homomorphism betweeen the two. If this is surjective, then if $S$ is a group, so is $T$. 

I don't understand what there is to prove. Isn't the definition of a homomorphism between two structures a map between the two structures? So if it's a homomorphism, if one is a ring, the other is a ring; if one is a field, the other is a field; etc.
 A: The only hypothesis we are given is that these are two structures with binary operations. The fact that there is a homomorphism between them a priori implies nothing without knowing something about the homomorphism.
Some examples:
There is a homomorphism from $\mathbb{Z}_4$ into the multiplicative monoid of the Gaussian integers $\mathbb{Z}[i]$ sending a generator to $i$. This is not a ring homomorphism, it is only a group homomorphism, and the multiplicative monoid of $\mathbb{Z}[i]$ is not a group, even if we exclude $0$, because some elements don't have inverses.
There is a homomorphism from $\mathbb{Q}$ into $\mathbb{Q}[x]$ given by $1\mapsto 1$. Even though this is a ring homomorphism and $\mathbb{Q}$ is a field, $\mathbb{Q}[x]$ is not a field.
There is a homomorphism from $\mathbb{Z}$ into any group by sending $1$ to any element and extending accordingly. Even though $\mathbb{Z}$ is a ring, the group may not be a ring. Indeed, it need not even be the additive group of a ring because it may not be abelian.
There are many other examples.
A: Let $\varphi:S\to T$ be your homomorphism. We need to check that $(T,\cdot)$ satisfies the axioms of a group.


*

*If $e$ is the identity of $S$, we need to show that $\varphi(e)$ is the identity of $T$.

*Since $\varphi(a)\varphi(a^{-1})=\varphi(e)$, we can show that every element of $T$ has an inverse ($\varphi$ is surjective).

*Associativity follows from the fact that
$$
(\varphi(a)\varphi(b))\varphi(c)=\varphi(ab)\varphi(c)=\varphi((ab)c)
$$
and $S$ has associativity.

