How many elements are in the kernel of the homomorphism $f:(\mathbb{Z}/154\mathbb{Z})^* \to (\mathbb{Z}/154\mathbb{Z})^* $ where $f(x)=x^5$? The group operation in this case is multiplication with identity element $1 \mod 154$, so the group we are considering is $((\mathbb{Z}/154\mathbb{Z})^*,\cdot,\overline{1})$. The kernel of a homomorphism is given by:
$$\operatorname{Ker}(f):=\{ g\in G|f(g) =e_G\}$$
There is a theorem that states that for any element in a group, the order of that element is either infinite or it must divide the number of elements in the group. In $(\mathbb{Z}/154 \mathbb{Z})^*$ there are 154 elements and $5 \nmid 154$. Therefore there are no element of order 5 and because 5 is prime no element other than the identity satisfies $x^5=e$. Therefore the kernel of $f$ contains only the identity element $1 \mod 154$. Is this correct or are there more elements in the kernel? The reason for this question is that I am not sure if this argument alone is enough. Thanks in advance!
EDIT I realise that I completely forgot to add the $^*$ to the group $\mathbb(Z)/154\mathbb{Z}$ when I first wrote this question and subsequently in trying to determine the number of elements made a mistake. I realise now that I need to evaluate $\phi(154)$ to find the number of elements and see if $5|\phi(154)$ where $\phi$ is the Euler-Totient function. I am sorry for the confusion.
Best Answer
The problem is that $\mathbb{Z}_{154}^\star $ does not, as you've said, have $154$ elements. Note that $\varphi(154)=60$.