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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.

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  • $\begingroup$ 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. $\endgroup$ – user296602 Feb 1 at 2:54
  • $\begingroup$ You are writing the group multiplicatively, not additively. You are adding integers $m$ and $n$. $\endgroup$ – J. W. Tanner Feb 1 at 2:57
  • $\begingroup$ How does your book define $g^n$ for $n \in \mathbb{Z}$? $\endgroup$ – Theo Bendit Feb 1 at 2:57
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    $\begingroup$ For (b), it suffices to show $(g^n)(g^{-n})$ is the identity element. $\endgroup$ – angryavian Feb 1 at 3:01
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    $\begingroup$ @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 $(g^{-1})^n$? $\endgroup$ – Theo Bendit Feb 1 at 3:05
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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.

As this looks like homework, I'll give a hint for each question.

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 $G$?

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