# Induction for all integers

• Mr Davis 97
In summary: Because you used the fact, that ##\mathbb{Z}## is symmetric relative to the sign. You have proven the statement for all positive integers and zero, and then transformed this result to the negative integers by taking the power to ##-1##. In a way you used the same induction twice, or more precisely: you solved a portion of the statement and used another proof based on this result to do the rest.

## Homework Statement

Let ##\phi : G \to H## be a homomorphism. Prove that ##\phi (x^n) = \phi (x)^n## for all ##n \in \mathbb{Z}##

## The Attempt at a Solution

First, we note that ##\phi (x^0) = \phi(x)^0##. This is because ##1_G \cdot 1_G = 1_G \implies \phi (1_G 1_G) = \phi (1_G) \implies \phi (1_G)^2 = \phi (1_G) \implies \phi (1_G) = 1_H##.

Second, we show that ##\phi (x^n) = \phi (x)^n## where ##n \in \mathbb{Z}^+##. The base case is trivial. Now, suppose for some ##k## we have that ##\phi (x^k) = \phi (x)^k##. Then ##\phi (x^{k+1}) = \phi (x^k x) = \phi (x^k) \phi (x) = \phi (x)^k \phi (x) = \phi (x)^{k+1}##. This proves the result for positive integers.

Third, we prove the result for negative integers. First, we show that ##\phi (x^{-1}) = \phi (x)^{-1}##. ##xx^{-1} = 1_G \implies \phi (xx^{-1}) = 1_H \implies \phi(x) \phi(x^{-1}) = 1_H \implies \phi (x^{-1}) = \phi(x)^{-1}##. So, since the result ##\phi (x^n) = \phi (x)^n## is true for all positive integers, the result ##\phi (x^{-n}) = = \phi ((x^{-1})^n) = \phi (x^{-1})^n = ( \phi (x)^{-1} )^n = \phi (x)^{-n}## must also be true for all positive integers.

Is this argument fine?

• David Dyer
Mr Davis 97 said:
Is this argument fine?
Apart from a double "=", yes. Maybe you should have mentioned ##x^0=1_G## and ##\phi(x)^0=1_H## but this is nit-picking. Your proof is fine.

• Mr Davis 97
fresh_42 said:
Apart from a double "=", yes. Maybe you should have mentioned ##x^0=1_G## and ##\phi(x)^0=1_H## but this is nit-picking. Your proof is fine.
Great, but I have a couple of questions. So, I proved that ##\phi (x^n) = \phi (x)^n##. Why was it valid for me to substitute ##x## with ##x^{-1}## to get the result ##\phi (x^{-n}) = \phi (x)^{-n}##? Why didn't I have to do an induction a second time for the negative case?

Mr Davis 97 said:
Great, but I have a couple of questions. So, I proved that ##\phi (x^n) = \phi (x)^n##. Why was it valid for me to substitute ##x## with ##x^{-1}## to get the result ##\phi (x^{-n}) = \phi (x)^{-n}##? Why didn't I have to do an induction a second time for the negative case?
Because you used the fact, that ##\mathbb{Z}## is symmetric relative to the sign. You have proven the statement for all positive integers and zero, and then transformed this result to the negative integers by taking the power to ##-1##. In a way you used the same induction twice, or more precisely: you solved a portion of the statement and used another proof based on this result to do the rest.

• Mr Davis 97