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GRE Subject Test Question:

Up to isomorphism, how many abelian groups are there of order 36?

The answer given is 4 and the explanation is as follows: Let G be an abelian group with order n. Then G is isomorphic to the products of the form $$Z_{(p_1^{n_1})} \times Z_{(p_2^{n_2})} \times \cdots \times Z_{(p_k^{n_k})}$$ where the $p_j$'s are not necessarily distinct, are the primes in the factorization of n and that $(p_1^{n_1})(p_2^{n_2}) \cdots (p_k^{n_k})=n$. Here $Z_n$ denotes the cyclic group of $\{0,1,2,3,4,5,6,7\}$ under addition modulo n. For $n = 36 = 2^2 3^2$, G is isomorphic to $$Z_{(2^2)} \times Z_{(3^2)} \\ Z_{(2^1)} \times Z_{(2^1)} \times Z_{(3^2)} \\ Z_{(2^2)} \times Z_{(3^1)} \times Z_{(3^1)} \\ Z_{(2^1)} \times Z_{(2^1)} \times Z_{(3^1)} \times Z_{(3^1)}$$

I have two questions:

  1. Could someone provide me with a link to a proof of the statement that G is isomorphic to the products of the form $Z_{(p_1^{n_1})} \times Z_{(p_2^{n_2})} \times \cdots \times Z_{(p_k^{n_k})}$ ?
  2. Why does $Z_n$ denote the cyclic group of $\{0,1,2,3,4,5,6,7\}$ under addition modulo n?

Thanks in advance and sorry if this is a basic question. I'm a physics major trying to make the transition to applied mathematics in graduate school and I'm afraid my abstract mathematics are rather weak!

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    $\begingroup$ $Z_n $ denote {$0,1,2,3....,n-1$}, that is $Z_8$ what you have written $\endgroup$ – Bhaskar Vashishth Oct 2 '14 at 22:18
  • $\begingroup$ @BhaskarVashishth Thank you so much for the link, it's exactly what I was looking for. Regarding your second comment, that is what I thought. $Z_n$ refers to the cyclic group of integers under addition modulo n. So when the book explanation says that $Z_n$ denotes the cyclic group of $\{0,1,2,3,4,5,6,7\}$ under addition modulo n, would you agree that it's a misprint? I was having trouble understanding why the set of integers would stop at 7 when n could in fact be much larger than 7. $\endgroup$ – Blaize Berry Oct 3 '14 at 1:02
  • $\begingroup$ yaeh or may be he was taking $n=7$ $\endgroup$ – Bhaskar Vashishth Oct 3 '14 at 1:08
  • $\begingroup$ Thanks so much for the clarification. Really appreciate your help! $\endgroup$ – Blaize Berry Oct 3 '14 at 1:11

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