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Let $G$ be the group of units of the commutative group $\mathbb{Z}/2007\mathbb{Z}$ and consider the group homomorphism $h\colon G\to G$ where $h(g)=g^n$ for $n\ge1$.

How can I find the order of $\ker h$?

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The kernel of $h$ is the elements satisfying $g^n=1$. Do you know how to find the number of elements of a given order in a cyclic group? Somewhere along the way, you will have to also find the factorization of 2007.

EDIT: By the way, I'm a little worried about your phrase, "the group of units of the commutative group ${\bf Z}/2007{\bf Z}$." ${\bf Z}/2007{\bf Z}$ is a commutative group under addition and, as a group, all its elements are units. But it seems that what you really mean is "the group of units of the commutative ring ${\bf Z}/2007{\bf Z}$." Those would be the elements that have multiplicative inverses, and that is what we are discussing in the comments.

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I know $2007=3\cdot 3\cdot 223$. So by the Chinese remainder theorem, $G$ is isomorphic to $(\mathbb{Z}/9\mathbb{Z})^*\times(\mathbb{Z}/223\mathbb{Z})^*$. The order of these two group is easy to find, since $$(\mathbb{Z}/223\mathbb{Z})^*$ is of order 222 and $(\mathbb{Z}/9\mathbb{Z})^*$ is of order 8. But I don't know how to go on from here. – Mark Jun 25 '12 at 13:30
Actually, there are only 6 units in ${\bf Z}/9{\bf Z}$, not 8. Indeed, ${\bf Z}/9{\bf Z}$ is cyclic of order 6, and the other bit is cyclic of order 222. So you need to work out how many elements of each possible order in what I'll write as $C_6\oplus C_{222}$. The only possible orders are the divisors of 222, namely, 1, 2, 3, 6, 37, 74, 111, and 222. Can you work out how many elements of each order? – Gerry Myerson Jun 25 '12 at 13:40

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