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Let $G=\mathbb{Z}_4$, the group of integers modulo $4$, and let $H$ be the Klein four group, let $f: G \rightarrow H$ be a homomorphism. Why does the kernel of $f$ contain the element $2$ of $G$?

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2 Answers 2

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Let's find the possible homomorphisms.

First we require $f(0) = 1_H$

So now we need $1_H = f(0) = f(1+3) = f(1)f(3)$. So $f(1) = f(3)$.

Now $1_H = f(3)^2 = f(3 + 3) = f(2) = f(1+1) = f(1)^2 = 1_H$.

So all homomorphisms are of the form $\{0, 1, 2, 3\} \mapsto \{1_H, h, 1_H, h\}$ where $h$ is any element in the Klein 4 group.

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Using the homomorphism property, $$f(2)=f(1+1)=f(1)^2.$$ What is $f(1)^2$ in the Klein 4 group? (Try out all the different possible values for $f(1)$ if you're not sure.)

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  • $\begingroup$ f(1)^2=1 and its in the kernel of f? $\endgroup$
    – Yang Ethan
    May 27, 2016 at 3:00
  • $\begingroup$ Yep, you've got it. $\endgroup$
    – kccu
    May 27, 2016 at 3:16

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