What is the smallest positive integer n such that there are exactly four non-isomorphic abelian groups of order n?

This is a question in Joseph A.Gallian's book, and the answer is $n=36$ and the groups are : $\mathbb Z_9× \mathbb Z_4$; $\mathbb Z_3×\mathbb Z_3×\mathbb Z_4$; $\mathbb Z_9×\mathbb Z_2×\mathbb Z_2$; $\mathbb Z_3×\mathbb Z_3×\mathbb Z_2×\mathbb Z_2$;

Now my question is when $n=16$ then the group is isomorphic to Z(16) and hence abelian.

And $\mathbb Z_{16}$ is of order $16$ so there are $4$ non isomorphic abelian groups:

$\mathbb Z_{16}$;

$\mathbb Z_8×\mathbb Z_2$;

$\mathbb Z_4×\mathbb Z_2×\mathbb Z_2$;

$\mathbb Z_2×\mathbb Z_2×\mathbb Z_2 ×\mathbb Z_2$;

What is wrong in my answer?

  • 1
    $\begingroup$ The frase the answer of @Theophile more easily (in my opinion), you missed Z(4)xZ(4) in your answer, so there are 5 non-isomorphic abelian groups of order 16, not 4. $\endgroup$
    – Krijn
    Jun 7 '15 at 14:36

The number of non-isomorphic Abelian groups depends on the partition number of the exponents in the prime factorization. Let $P(k)$ be the number of ways to partition $k$ (for example, $P(4) = 5$, since we have $4, 3+1, 2+2, 2+1+1$, and $1+1+1+1$). Then for $n=p_1^{e_1} \ldots p_r^{e_r}$, there are $$\prod_i P(e_i)$$ non-isomorphic Abelian groups of order $n$. You missed $2+2$ among the ways to partition $4$, corresponding to $\mathbb Z/(2^2\mathbb Z) \times \mathbb Z/(2^2\mathbb Z) = \mathbb Z/4\mathbb Z \times \mathbb Z/4\mathbb Z$.

You can use this formula to answer the original question. The first few partition numbers are:

P(1) = 1    1
P(2) = 2    2, 1+1
P(3) = 3    3, 2+1, 1+1+1
P(4) = 5    4, 3+1, 2+2, 2+1+1, 1+1+1+1
P(5) = 7    5, 4+1, 3+2, 3+1+1, 2+2+1, 2+1+1+1, 1+1+1+1+1

and they obviously increase from here. Now, if there are $4$ Abelian groups of order $n$, then we must have $e_1=e_2=2$, and we choose as bases the two smallest primes. Thus the number we are looking for is $2^23^2=36$.


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