abstract algebra: cyclic!

**dlin3** · September 19th 2008, 01:46 PM

can someone please help? i'm studying for an exam and i'm stuck on this problem.

Prove that U(2^n) (n>less than 3) is not cyclic.

thanks so much

**ThePerfectHacker** · September 19th 2008, 02:02 PM

Originally Posted by dlin3

can someone please help? i'm studying for an exam and i'm stuck on this problem.

Prove that U(2^n) (n>less than 3) is not cyclic.

thanks so much

1)Prove that $5^{2^{n-3}} \equiv 1 + 2^{n-1} ~ (2^n)$ for $n\geq 3$ by induction.
2)Now show that $5^{2^{n-2}} \equiv 1 ~ (2^n)$ and $5^{2^{n-3}} \not \equiv 1 ~ (2^n)$ .
3)Show that $\{ \pm 5^k | 0\leq k < 2^{n-1} \} = U(2^n)$ .
4)Therefore no element has order $\phi (2^n)$ in $U(2^n)$ .

**dlin3** · September 22nd 2008, 02:29 PM

How did you get 5^(2n-3) = 1 + 2^(n-1) * (2n)?

**ThePerfectHacker** · September 22nd 2008, 04:53 PM

Originally Posted by dlin3

How did you get 5^(2n-3) = 1 + 2^(n-1) * (2n)?

Use what I said above. Use induction.
And the fact if $a\equiv b(\bmod p^k)$ then $a^p \equiv b^p (\bmod p^{k+1})$ (where $p$ is prime, in this case two).

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