Show that the following map is a group homomorphism and find its kernel. State whether the mapping is injective or surjective.

$\phi : \mathbb{Z} \to (\{1,-1\},{\times})$ by $\phi(a) = (-1)^{a}$

$\phi(a+b) = (-1)^{(a+b)} = (-1)^{a} (-1)^{b} = \phi(a) \phi(b) $

I do not understand why the book says the kernel is $= 2\mathbb{Z}$?

How is this the identity element in the mapping?

And how to identify whether this is surjective or injective?

  • $\begingroup$ A hint for the injective/surjective part: You likely have a theorem about how the kernel of a homomorphism relates to injectivity. $\endgroup$
    – anakhro
    Jun 16 '15 at 13:54

The identity element in the group $\{\pm 1\}$ is $1$. The kernel consists of all the elements that $\phi$ sends to the identity element $1$. That is, $$ \ker \phi = \{n\in \mathbb{Z} : (-1)^n = 1\} = ... $$ (So you just need to solve $(-1)^n = 1$.)

Is the map surjective? Well, can you get both $1$ and $-1$?

Is the map injective? The map is not injective if you can find two different numbers $n$ and $m$ such that $(-1)^n = (-1)^m$. Can you do that?


The kernel is the set of elements $a$ such that $\phi(a) = (-1)^a = 1$. This clearly holds if and only if $a$ is even, and so $\ker(\phi) = 2\mathbb Z$.

The map is injective if and only if its kernel is $\{0\}$, which we've just shown is not true. The map is surjective if it "hits" every element; clearly, $\phi(0) = 1$ and $\phi(1) = -1$, so the map is surjective.

  • $\begingroup$ Oh I see it now. thank you, makes a lot of sense now! $\endgroup$
    – italy
    Jun 16 '15 at 13:56

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