Subgroups of $G$ if there is a homomorphism from $G$ to $C_2$

Question

Is there always a subgroup of index 2 in $G$ if there is a homomorphism $f: G \rightarrow C_2$? My inkling is that this is true because in this case $G$ must have a normal subgroup of index 2, and the result follows from Lagrange's theorem. Could someone please help me and provide a solution?

Answer 1

Need not have a subgroup of index 2.

For counter-example: It is well known that $A_5$ does not have a subgroup of index 2. But, there is a trivial homomorphism from $A_5$ to $C_2$. Even if it is not onto, which is not mandatory in the question.

Answer 2

Just to elaborate on the other answer and comments:

Given any two groups $G$ and $H$, there is always a homomorphism from $G$ to $H$, namely the trivial map which sends every element of $G$ to the identity in $H$.

What you want is the following:

Theorem.

$G$ has a subgroup of index $2$ if and only if there is a non-trivial homomorphism $G \to C_2$.

Proof.

Note first that because $C_2$ has only $2$ elements, a homomorphism $G \to C_2$ is surjective if and only if it is non-trivial. Now:

• ($\Rightarrow$). If $G$ has a subgroup $N$ of index $2$, this subgroup is automatically normal and therefore $G / N$ is a group which has order $2$ and hence is isomorphic to $C_2$. Therefore the projection map $G \to G/N$ gives a surjective homomorphism from $G$ to $C_2$ which as mentioned above, must be non-trivial.
• ($\Leftarrow$). If there exists a non-trivial homomorphism $\varphi: G \to C_2$, as mentioned above, $\varphi$ must be surjective. Hence by Lagrange, the kernel of $\varphi$ has index $2$.