# If $G$ is a group and $g\in G$, then for all $n,m\in\mathbb{Z}$, we have the following properties.

Proof attempt:

(a) $$g^ng^m=g^{n+m}$$

(b) $$(g^n)^{-1}=g^{-n}$$

Proof(informal rough draft).

(a) Since $$g\in G$$, we can rewrite $$g^n=gg...g$$ for n-factors of $$g$$ and $$g^m=gg...g$$ for m-factors of $$g$$, since rules of exponents still hold for groups(given definition in the book). If $$G$$ is under a group under addition, we can add the exponents to obtain $$g^ng^m=g^{n+m}$$.

Have not yet began (b).

Thoughts? This is an intro to group theory/proof class so instructor said it should be simple yet very direct.

• If the basis of your proof is "rules of exponents still hold for groups," then you're just assuming exactly what you're trying to prove... so I think you're oversimplifying it. A better way is to do a quick induction argument. – user296602 Feb 1 at 2:54
• You are writing the group multiplicatively, not additively. You are adding integers $m$ and $n$. – J. W. Tanner Feb 1 at 2:57
• How does your book define $g^n$ for $n \in \mathbb{Z}$? – Theo Bendit Feb 1 at 2:57
• For (b), it suffices to show $(g^n)(g^{-n})$ is the identity element. – angryavian Feb 1 at 3:01
• @Ryan Ugh. I much prefer a recursive definition, as it's more precise and easier to work with in proofs. But, also importantly, how do they define $g^{-n}$ for $n > 0$? Is it just $(g^{-1})^n$? – Theo Bendit Feb 1 at 3:05

Your proof of part (a) is, as mentioned in the comments, not a proof; you are assuming that which you wish to show. Furthermore, writing "If $$G$$ is under a group [sic] under addition, we can add the exponents to obtain $$g^n g^m = g^{n+m}$$ is not just assuming that which you want to show, but also is misleading: this identity holds in any group, and the question asks you to show that it indeed holds for any group, not just a group with addition as its operation.
For part (a), you say that the definition of $$g^n$$ that you were given is as the product of $$n$$ copies of $$g$$. What happens if you multiply $$n$$ copies of $$g$$, and then multiply this by $$m$$ copies of $$g$$? How do you write this product in your group, before and after you carry out the multiplication?
For part (b), as stated in the comments, it suffices to show that $$g^n g^{-n}$$ is the identity element. As (presumably, analogous to the definition in (a)) the definition of $$g^{-n}$$ given is the product of $$n$$ copies of $$g^{-1}$$, what would you get if you first multiply $$n$$ copies of $$g$$, followed by $$n$$ copies of $$g^{-1}$$? How would you write either product in $$G$$?