Proving Induction for All Integers

[Mr Davis 97]

Homework Statement

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

Homework Equations

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?

 

[fresh_42]

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]

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?

 

[fresh_42]

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.