If $f:G→ℤ_{8}$ be a group epimorphism , then what can be said about $G$?

Question

I was trying to answer the following problem :

Let $G$ be a finite group and $ f : G \to \Bbb Z_8$ be a group epimorphism, then which of the following must be true ?

(a) $G$ is isomorphic to $Z_8$ .

(b) $G$ has a subgroup of order 8 .

(c) $G$ has a normal subgroup of order 8 .

(d) $G$ has a normal subgroup of index 2 .

My attempt:

Option (a) is clearly not true, since we can choose our $G$ to be $Z_{16}$ and $f$ to be defined as follows : $ f :Z_{16} \to Z_{8}$ , $ f(\bar 1) = \bar 1$ .

But I'm unable to draw any other conclusion. Thanks in advance for help.

Answer 1

Assertion (d) holds: take $H=f^{-1}\bigl(\{0,2,4,6\}\bigr)$. Then, since $\{0,2,4,6\}$ is a normal subgroup of $\mathbb{Z}_8$, $H$ is a normal subgroup of $G$. Furthermore$$G/H\simeq\mathbb{Z}_8/\{0,2,4,6\}\simeq\mathbb{Z}_2.$$

Answer 2

The other answers show that (d) is true. Also the hypothesis implies that $8$ divides $|G|$, so (b) is true. But (c) is not always true. An example is a wreath product of $C_3$ with $C_8$ - or any group with more than one Sylow $2$-subgroup that is isomorphic to $C_8$.