1

I'm posting this question because user last seen is more than $5$ years ago

I have some confusion on this answer

It is written that

Let $G$ be the set of non-zero elements of $\mathbb{Z}_p$. $G$ is a commutative group of order $p-1$. Let $f\colon G \rightarrow G^2$ be the map such that $f(m) = m^2$. $f$ is a surjective homomorphism. Hence $G^2$ is isomorphic to $G/\ker(f)$. Since $\ker(f) = \{1, -1\}$ and $p$ is odd, $|ker(f)| = 2$. Hence $|G^2| = (p-1)/2$.

My question : Why is $f$ a surjective homomorphism ?

My thinking :Let define the map $f : \mathbb{Z}^*_p \to \mathbb{Z}^*_p $ by $f(x)=x^2$

Here order of $ G=\mathbb{Z}^*_p= p-1$ where $p$ is odd prime

We can't use Bezout 's theorem here because $2$ and $p-1$ are not relatively prime

This implies $f$ is not surjective

For example take $G= \mathbb{Z}^*_3$

$\mathbb{Z}^*_p = \{1,2,3,...,p-1\} \implies \mathbb{Z}^*_3=\{1,2\}$

$f:\mathbb{Z_3}^* \to \mathbb{Z_3}^*$ defined by $f(x) =x^2$

$f(1)=1$ , $f(2)=2^2=4=1 \bmod 3$

Therefore $f$ is not onto

Arturo Magidin
  • 398,050
user12392
  • 183
  • 5
    Note that the range in the quote is $G^2$: the set of all squares of elements of $G$. So, by definition, the map is surjective. Your thinking uses the wrong codomain, it should be $(\mathbb{Z}_p^)^2 = {x^2\mid x\in\mathbb{Z}_p^}$, which for $p=3$, would be ${1^2,2^2} = {1}$. – Arturo Magidin Mar 28 '22 at 19:39
  • @ArturoMagidin $G^2 = \mathbb{Z}^_p \times \mathbb{Z}^_p $.Im not getting why is $f : \mathbb{Z}^_p \to \mathbb{Z}^_p \times \mathbb{Z}^*_p $ onto? – user12392 Mar 28 '22 at 19:43
  • 2
    No, here $G^2$ is being used in the sense of group theory, not set theory; for a multiplicative group $G$, and positive integer $n$, $G^n = \langle x^n\mid x\in G\rangle$; if the group is abelian, this equals ${x^n\mid x\in G}$. If the desired interpretation were $\mathbb{Z}^_p\times\mathbb{Z}^_p$, then the map would not even make sense, because $x^2$ is not an element of the cartesian product. – Arturo Magidin Mar 28 '22 at 19:45
  • okay @ArturoMagidin – user12392 Mar 28 '22 at 19:45
  • thanks you @ArturoMagidin – user12392 Mar 28 '22 at 19:45

0 Answers0