Mathematics Asked by James Eade on December 21, 2021
So I am missing something very basic here, but here it goes. As a consequence of the definition of a homomorphism we must have that $f(e) = e’$. Consider $(mathbb{Z}/5mathbb{Z})^*$ and the map $x to 2x$. It is a bijection but does not satisfies the multiplicative property but how can we claim it is an automorphism since $f(1) = 2 neq 1$.
Edit: Does not satisfy multiplicative property.
You're right that this is not a homomorphism! Indeed, it is not multiplicative: if $f(x) = 2x$ then $f(xy) = 2xy$ but $f(x)f(y) = (2x)(2y) = 4xy$. Then pick $x = y = 1$ to see that $f(xy) neq f(x)f(y)$.
Like you said, any multiplicative function on this set (a.k.a. any endomorpism of $mathbb{Z}_5^*$) will automatically send $1$ to $1$. Math is not broken!
Answered by diracdeltafunk on December 21, 2021
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