Prove that ord(a) = ord(bab^(-1))

**Iceflash234** · March 30th 2010, 09:25 AM

Prove that $ord(a) = ord(bab^{-1})$

Here's what I have so far:

Part 1: Show that $ord(bab^{-1}) = n$

Suppose that ord (a) = n,

Then $a^n = e$

The problem I'm having is using the Associative Law to show that $(bab^{-1})^n$ is also e.

Part 2: Show that ord (a) = n.

Suppose that $ord(bab^{-1}) = n$

Then $(bab^{-1})^n = e$

The problem I'm having at this part is using the Associative Law to show that $a^n = e$ .

Help is appreciated.

**Drexel28** · March 30th 2010, 09:31 AM

Originally Posted by Iceflash234

Prove that $ord(a) = ord(bab^{-1})$

Here's what I have so far:

Part 1: Show that $ord(bab^{-1}) = n$

Suppose that ord (a) = n,

Then $a^n = e$

The problem I'm having is using the Associative Law to show that $(bab^{-1})^n$ is also e.

Part 2: Show that ord (a) = n.

Suppose that $ord(bab^{-1}) = n$

Then $(bab^{-1})^n = e$

The problem I'm having at this part is using the Associative Law to show that $a^n = e$ .

Help is appreciated.

Well, clearly $(bab^{-1})=(bab^{-1}(bab^{-1})\cdots=ba^nb^{-1}=beb^{-1}=bb^{-1}=e$ so $|bab^{-1}|\leqslant n$ but what would happen if that inequality was strict?

**tonio** · March 30th 2010, 09:34 AM

Originally Posted by Iceflash234

Prove that $ord(a) = ord(bab^{-1})$

Here's what I have so far:

Part 1: Show that $ord(bab^{-1}) = n$

Suppose that ord (a) = n,

Then $a^n = e$

The problem I'm having is using the Associative Law to show that $(bab^{-1})^n$ is also e.

Part 2: Show that ord (a) = n.

Suppose that $ord(bab^{-1}) = n$

Then $(bab^{-1})^n = e$

The problem I'm having at this part is using the Associative Law to show that $a^n = e$ .

Help is appreciated.

Hint: $\forall\,\,n\in\mathbb{Z}\,,\,\,(bab^{-1})^n=ba^nb^{-1}$

And remember that it is NOT enough to show $x^k=e$ to conclude that $ord(x)=k$ : you must also show this k is the minimal natural number fulfilling that.

Tonio

**Iceflash234** · March 30th 2010, 10:08 AM

Thanks! That helped.

I see clearly how to show bab^{-1} = e.

I'm not sure though how to show a^{n} = e for the second part.

**proscientia** · March 31st 2010, 08:50 AM

Originally Posted by Iceflash234

Prove that $ord(a) = ord(bab^{-1})$

Here's what I have so far:

Part 1: Show that $ord(bab^{-1}) = n$

Suppose that ord (a) = n,

Then $a^n = e$

The problem I'm having is using the Associative Law to show that $(bab^{-1})^n$ is also e.

Part 2: Show that ord (a) = n.

Suppose that $ord(bab^{-1}) = n$

Then $(bab^{-1})^n = e$

The problem I'm having at this part is using the Associative Law to show that $a^n = e$ .

Help is appreciated.

You are way to confused there. What you need to show that is that $a^n=e\,\Leftrightarrow\,(bab^{-1})^n=e$ (which will imply that $\mathrm{ord}(a)=\mathrm{ord}(bab^{-1})).$ To do this, you need to use tonio’s hint (which can be easily proved by induction).

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