Thread: Order of elements in a group

1. Order of elements in a group

Prove that in any group, an element and its inverse has the same order.

My Proof:

Let G be a group, and suppose that $\displaystyle a \in G$ and $\displaystyle a^{-1} \in G$.

By definitions, $\displaystyle <a> = n$ such that $\displaystyle a^{n} = e$ and $\displaystyle <a^{-1} = m$ such that $\displaystyle (a^{-1})^{m} = e$

thus, $\displaystyle a^{n} = a^{-m}$

Now, can I say that n must equal to m because the negative of inverse equals to the original power?

2. Let $\displaystyle a\in G$ have finite order $\displaystyle n$, i.e. $\displaystyle a^n = 1$ and $\displaystyle n$ is the smallest such exponents.
Thus,
$\displaystyle \underbrace{a\cdot ... \cdot a}_n=1$
Thus,
$\displaystyle (a\cdot ... \cdot a)^{-1} = 1$
Thus,
$\displaystyle a^{-1} \cdot ... \cdot a^{-1} = 1$
Thus,
$\displaystyle (a^{-1})^n = 1$.
Now it remains to show that $\displaystyle n$ is smallest exponent for $\displaystyle a^{-1}$. Try showing that.

3. Now, $\displaystyle \underbrace{a^{-1} \cdot ... \cdot a^{-1}}_n=(a^{-1})^{n}=a^{-n}=(a^{n})^{-1}=(1)^{-1}=1$

Thus shows that $\displaystyle (a^{-1})^{n}=e=a^{n}$

Q.E.D.

Is that right?