I am trying to prove this following statement about groups.
Let $G$ be a group, and $a, b \in \mathbb{Z}$. Let $x \in G$.
(a) Show $(x^a)^{-1} = x^{-a}$.
(b) Show $x^{a+b} = x^a x^b$ and $(x^a)^b = x^{ab}$.
(c) Show that $|x| = |x^{-1}|$.
I am having difficulty proving these sequentially. The proof of (a) I would write, for example, would be $$x^a x^{-a} = x^{a + (-a)} = x^0 = e,$$ where $e$ is the identity in $G$. Similarly, $$x^{-a} x^a = x^{-a + a} = x^0 = e.$$ Hence, by uniqueness of the inverse in a group, $$(x^a)^{-1} = x^{-a}.$$ The problem is that I have used part (b) to prove part (a), and this is almost surely not the way I was supposed to do it. Without using (b), I cannot think of a way to prove (a). A hint would be very much appreciated.
I similarly cannot figure out how to prove (b), as this is a fact I have always assumed to hold and used as an auxilary lemma in proofs of this kind.
I believe I can prove see. I assume that writing $|x|$ implies that the order is finite. Otherwise, if $x$ has infinite order, then certainly $x^{-1}$ has infinite order, because $x^n \neq e$ for any $n$, so $x^{-n} \neq e$ for any $n$ because we can easily reverse engineer a statement about the order of $x$ or $x^{-1}$ from the other. Suppose that $|x|$ is finite and call it $m$. Then $$x^m = e.$$ But, using (a), $$(x^m)^{-1} = x^{-m}.$$ But $x^m = e$, and $ee = e$, so $(x^m)^{-1} = e$. That is, $$x^{-m} = e.$$ But we can rewrite the LHS: $$(x^{-1})^m = e.$$ Hence, $|x^{-1}| = m = |x|$.
Some help, especially on (a) and (b), would be appreciated.
It is possible to prove (a) independently. Some if not most textbooks have the standard notation (see Dummit and Foote for example) of \begin{equation*} a^{-n}=\underbrace{a^{-1}a^{-1}\cdots a^{-1}}_{n\text{ times}} \end{equation*} So with the above notation and first assuming $n>0$, \begin{equation*} a^{n}a^{-n}=\underbrace{aa\cdots a}_{n\text{ times}}\cdot\underbrace{a^{-1}a^{-1}\cdots a^{-1}}_{n\text{ times}}=1_{G} \end{equation*} where the last equality follows from the generalised associative law for groups. Therefore, $a^{-n}=(a^{n})^{-1}$. If $n<0$, then write $m=-n>0$ and use a similar argument as above.
To prove (b), let us consider (for simplicity sake) only the case when $b<0<a$ and $|b|<a$ in $\mathbb{Z}$. Once again, with the above notation and using the generalised associative law for groups, \begin{equation*} x^{a}x^{b}=x^{a}x^{-|b|}=\underbrace{xx\cdots x}_{a\text{ times}}\cdot\underbrace{x^{-1}x^{-1}\cdots x^{-1}}_{|b|\text{ times}}=x^{a-|b|}=x^{a+b}. \end{equation*} The other cases for (b) on $x^{a+b}=x^{a}x^{b}$ is done in a similar manner. To prove $(x^{a})^{b}=x^{ab}$, we once again consider only the case when $a>0$, $b<0$. In this case it does not matter if $|b|<a$ or $|b|>a$. Using the notation as above again, and using the result in (a) and the first part of (b), \begin{equation*} (x^{a})^{b}=(x^{a})^{-|b|}=((x^{a})^{-1})^{|b|}=(x^{-a})^{|b|}=\underbrace{x^{-a}x^{-a}\cdots x^{-a}}_{|b|\text{ times}}=x^{-a|b|}=x^{ab}. \end{equation*} The case when $a<0,b<0$ is done in a similar manner.