Group homomorphism that maps to one element Is there some way to construct a group homomorphism $G \to H$ that maps everything in $G$ to just one non-identity element of $H$ (besides mapping $0$ to $0$)? For arbitrary (finite) $H$ but you have your choice of $G$?
 A: The only example of this is the identity homomorphism $C_2\rightarrow C_2$ (where $C_2$ is the cyclic group of order $2$). Otherwise, if the image of a homomorphism consists of just one non-identity element and the identity, there are multiple elements mapping to the identity.
In particular, let $f:G\rightarrow H$ be a homomorphism. Then, it must hold that, for any $x\in H$ (and letting $1$ be the identity of $H$) such that $f^{-1}(x)$ is not empty:
$$|f^{-1}(x)|=|f^{-1}(1)|$$
- that is, the set of elements of $G$ mapping to $x$ is as large as the set of elements mapping to $1$ under any homomorphism. This is simple to prove, since, if we let $z$ be some element in the preimage $f^{-1}(x^{-1})$ - that is, some element satisfying $f(z)=x^{-1}$ - then it holds that
$$z\cdot f^{-1}(x)=f^{-1}(1)$$
where $z\cdot f^{-1}(x)$ is a coset of $f^{-1}(x)$. This can be proven since, if we apply $f$ to the left side, we would conclude that $f(z\cdot f^{-1}(x))=f(z)x=x^{-1}x=1$ - meaning every element of $z\cdot f^{-1}(x)$ is in $f^{-1}(1)$. Proving the converse - that every element in $f^{-1}(1)$ is in $z\cdot f^{-1}(x)$ is a simple exercise which I will leave to you.
However, what we can prove from this is that if the image of a homomorphism $f:G\rightarrow H$ consists of just two elements, then exactly half of the elements of $G$ map to the identity, and half the elements of $G$ map to the other element. If only one element is permitted to map to the identity, this gives that $1$ is half the order of $G$ - hence $G$ has order two.
A: The identity element of $G$ must map to the identity element of $H$.
Proof: Call our homomorphism $f: G \rightarrow H$. Then $f(x) = f(1_G \cdot x) = f(1_G) \cdot f(x)$. Now use cancellation: $f(x) \cdot f(x)^{-1} = f(1_G) \cdot f(x) \cdot f(x)^{-1}$. The LHS is $1_H$ and the RHS is $f(1_G)$.

Sorry, I seem to have misread. You want all non-identity elements of $G$ to map to the same element.
The only case is $G = C_2$ and the non-identity element of $H$ is of order two. The sizes of cosets of the kernel of the map must all be equal, so there must be an equal number of elements sent to the identity and to the non-identity element.
If you're willing to allow non-identity elements to map to the identity, then you need an order two element of $H$ (for it times itself must be in the set of itself and the identity), and many groups $G$ will admit a nonzero morphism to $C_2$.
A: Let such a homomorphism $f$ exist and all the non identity elements of $G$ are mapped to the non -identity element $b$ of $H$ .then Im$f$ will consist of $b$ and $e_H$ and ker$f$ will consist only of $e_G$.then by 1st isomorphism theorem
$|G|/|kerf|=|Imf|$
i.e$|G|=|Imf|$ snce ker$f$ consists only of $e_G$
i.e. $|G|=2$ which is the only possibility
