- Thread starter
- #1

I have proven this for m = 2, by simple calculation of the conjugates. Now I need to prove that if this is true for m = n then it is true for m = n + 1.

I appreciate any help you can offer, thank you.

- Thread starter gucci
- Start date

- Thread starter
- #1

I have proven this for m = 2, by simple calculation of the conjugates. Now I need to prove that if this is true for m = n then it is true for m = n + 1.

I appreciate any help you can offer, thank you.

- Moderator
- #2

- Jan 26, 2012

- 995

If you showed that $\overline{g_1g_2}=\overline{g_1}\cdot \overline{g_2}$, and then assume that $\overline{g_1g_2\cdots g_n} = \overline{g_1}\cdot\overline{g_2}\cdots \overline{g_n}$, then it follows that

I have proven this for m = 2, by simple calculation of the conjugates. Now I need to prove that if this is true for m = n then it is true for m = n + 1.

I appreciate any help you can offer, thank you.

\[\overline{g_1g_2\cdots g_ng_{n+1}} = \overline{(g_1g_2\cdots g_n) g_{n+1}}\]

Let's define $g_1g_2\cdots g_n=z$. Then we're left with $\overline{z g_{n+1}}$, and we know from the $m=2$ case that this is the same as $\overline{z}\cdot\overline{g_{n+1}}$. Now rewrite this as $\overline{g_1g_2\cdots g_n}\cdot \overline{g_{n+1}}$ and apply the inductive hypothesis to finish the problem.

Does this make sense?

- Thread starter
- #3