Order of elements in a group

**tttcomrader** · August 28th 2007, 08:24 AM

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

My Proof:

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

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

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

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

**ThePerfectHacker** · August 28th 2007, 08:29 AM

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

**tttcomrader** · August 28th 2007, 09:20 AM

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

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

Q.E.D.

Is that right?

**ThePerfectHacker** · August 28th 2007, 01:48 PM

Originally Posted by tttcomrader

Is that right?

Almost. You need to show this is the smallest such exponent.

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