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I'm supposed to show that if $p$ is an odd prime (so its not $2$) that the multiplicative group $(\mathbb{Z}/p^{a}\mathbb{Z})^*$ is cyclic. It seems very hard to do for any $a$, so I'm thinking about using induction on a to make things easy. That being said, I don't really like group theory. How do I show the base case is cyclic and proceed?

Thanks everyone!

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The additive group is cyclic? –  EuYu Nov 6 '12 at 20:17
    
Why downvote?${}$ –  wj32 Nov 6 '12 at 20:26
    
Hey thanks for the response and for the edits! It loots great. I don't understand the your comment though @EuYu –  Chloe.H Nov 8 '12 at 2:25
    
The additive group implies that addition is your binary operation. I think you meant the multiplicative group. The additive group is simply $1,\ 1+1,\ 1+1+1,\ \cdots$ so it's trivially cyclic. –  EuYu Nov 8 '12 at 2:28
    
oh sorry! ya of course, thank you! –  Chloe.H Nov 8 '12 at 2:34

2 Answers 2

up vote 1 down vote accepted

For each $d|p-1$ let $X_d$ bet the set of units of $\mathbb Z/p\mathbb Z$ of order $d$, $(\mathbb Z/p\mathbb Z)^\times$ is the disjoint union of each $X_d$.

The elements of $X_d$ are roots of the cyclotomic polynomial $\Phi_d$ of degree $\varphi(d)$, in any field (in this case we use that $(\mathbb Z/p\mathbb Z)^\times$ is a field) a degree $d-1$ polynomial has $\le d-1$ roots.

Therefore $\sum_{d|p-1} \varphi(d) = p-1 = \sum_{d|p-1} |X_d|$ so $|X_d| = \varphi(d)$.

That shows that $(\mathbb Z/p\mathbb Z)^\times$ is cyclic, the next step is to prove $(\mathbb Z/p^2\mathbb Z)^\times$ cyclic by creating a generator for it using the generator of $(\mathbb Z/p\mathbb Z)^\times$. Then you can prove $(\mathbb Z/p^r\mathbb Z)^\times$ cyclic by showing that the generator for $(\mathbb Z/p^2\mathbb Z)^\times$ is also a generator for it.

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"Show (Z/p2Z)× is cyclic by showing that if g generates (Z/pZ)×, then g or g+p generates (Z/p2Z)× " is this how to do it? –  Chloe.H Nov 9 '12 at 0:15
    
@Chloe.H, yes . –  sperners lemma Nov 9 '12 at 14:19
    
@spernerslemma: $({\mathbf Z}/p{\mathbf Z})^\times$ is not a field. –  KCd Oct 20 '13 at 18:27

The post marked answer shows the base case and sets up the induction. Show $(Z/p^2Z)^\times$ is cyclic by showing that if $g$ generates $(Z/pZ)^\times$, then $g$ or $g+p$ generates $(Z/p^2Z)^\times)$.

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Ok I managed to do this. But how do I apply this inductively? –  Chloe.H Nov 9 '12 at 0:04
    
I dont think I can mark both of these answwers right sorry! –  Chloe.H Nov 9 '12 at 0:07
    
so g or g+p are generators for (Z/p^a Z)* ? –  Chloe.H Nov 9 '12 at 0:11

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