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Let $A$ and $B$ be finite abelian groups.

Suppose that for every natural number $m$, the number of elements of order $m$ in $A$ is equal to the number of elements of order $m$ in $B$.

Prove that $A$ and $B$ are isomorphic.


Idea

Given that these groups are finite, I think you have to use the primary decomposition theorem somehow.

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    $\begingroup$ Hello, welcome to Maths.SE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ – Lord_Farin Dec 29 '14 at 13:45
  • $\begingroup$ @Mathmo123 They do. Can you show me an element of finite order in $A$ that has a different order than an element of finite order in $B$? =) $\endgroup$ – Pedro Tamaroff Dec 29 '14 at 14:05
  • $\begingroup$ @PedroTamaroff ah of course, the obvious one. I stand corrected. $\endgroup$ – Mathmo123 Dec 29 '14 at 14:06
  • $\begingroup$ my mistake, it is given that theses groups are finite, i think you have to use the primary decomposition theorem somehow... $\endgroup$ – Tom Turner Dec 29 '14 at 14:29
  • $\begingroup$ Duplicate of math.stackexchange.com/questions/72944/… $\endgroup$ – Gerry Myerson May 16 '18 at 0:47
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Here's how I would look at it. Suppose you are given the order $g$ of the group which has canonical factorization $p_1^{\alpha_1} p_2^{\alpha_n} \dots p_n^{\alpha_n}$. By the fundamental theorem of finite group your group is direct product of $n$ abelian $p$-groups, which are in turn direct products of cyclic subgroups of different orders. We want to determine for each prime which $p$-group is being used.

To do this look at the elements whose order has only $p$ as a prime divisor. Those are the elements which are of the form $(1,1,1,g,1,1\dots 1)$ These elements form a subgroup which is isomorphic to the $p$-subgroup.

So we would like to discern which subgroup it is by knowing the elements of order power of $p$.

To prove we can do this we need to prove that if

$\mathbb Z_{p^{a_1}}\times\mathbb Z_{p^{a_2}}\dots \mathbb Z_{p^{a_r}}$ and $\mathbb Z_{p^{b_1}}\times\mathbb Z_{p^{b_2}}\dots \mathbb Z_{p^{b_s}}$ are groups of order $p^n$ with different exponents they have a different count of orders.

To see this note that the order of an element in a direct product $g=(g_1,g_2\dots g_n)$ is the least common multiple of all of the orders, in a $p$-group it is the highest order.

So go order the factors from least to greatest. and suppose they differ for the first time at factor $k$, where the first group has a larger exponent (call it $p^x$) Then that group has more elements of order $p^x$.

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  • $\begingroup$ i still do not understand why in the last claim, if the first group has a larger exponent(p^x) the group must have more elements of order p^x. $\endgroup$ – Tom Turner Dec 30 '14 at 18:45
  • $\begingroup$ Well, that isn't true by itself, the point is that an element of the direct product in a p-group has order $p^x$ if and only if the element that has the highest order inside the parenthesis has order $p^x$. We take the two p-groups and write them out in decreasing order of the exponenents and look at the first part where they are different ok? $\endgroup$ – Jorge Fernández Hidalgo Dec 30 '14 at 18:54
  • $\begingroup$ So call the product of the first factors that are equal $K$ and let the groups be $G$ and $H$. $\endgroup$ – Jorge Fernández Hidalgo Dec 30 '14 at 18:55
  • $\begingroup$ Then $G=K\times Z_{p^x}\times Z_{p^{g_1}}\dots$ and $H=K\times Z_{p^y}\times Z_{p^{h_1}}\dots$. We only care about $p^x$ and $p^y$, suppose $x>y$, then $G$ has more elements of order $x$, why? they both have the elements of order $p^x$ that have that order because of the factor coming from $K$, but $G$ has more, those afforder by having an element from $Z_{p^x}$ $\endgroup$ – Jorge Fernández Hidalgo Dec 30 '14 at 18:58
  • $\begingroup$ i get it now, thank you so much! $\endgroup$ – Tom Turner Dec 30 '14 at 19:25
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This is not true in general. Take $A = \mathbb{Q}, B = \mathbb{Q} \times \mathbb{Q}.$ Even if you assume that the groups contains "some" elements of finite order, then also it is not true. Take $A = \mathbb{Q} \times \mathbb{Z}/m\mathbb{Z}, B = \mathbb{Q} \times \mathbb{Q} \times \mathbb{Z}/m\mathbb{Z}$ where $m > 1$ is an integer.

But if you choose $A$ and $B$ to be finite, then it is true. This follows from the structure theorem of finite abelian groups. Note also that it is not true even for finitely generated abelian groups which are not finite. Take $A = \mathbb{Z}^r, B = \mathbb{Z}^s, r\neq s.$

$\bf{EDIT:}$ Let $A$ and $B$ be two finite abelian groups. Then $A \cong \mathbb{Z}/d_1\mathbb{Z} \times \mathbb{Z}/d_2\mathbb{Z} \times \cdots \times \mathbb{Z}/d_r\mathbb{Z},$ for some $d_1, d_2, \cdots ,d_r \in \mathbb{Z}$ are prime powers (not necessarily distinct). Similarly $B \cong \mathbb{Z}/e_1\mathbb{Z} \times \mathbb{Z}/e_2\mathbb{Z} \times \cdots \times \mathbb{Z}/e_s\mathbb{Z},$ for some $e_1, e_2, \cdots , e_s \in \mathbb{Z}$ are prime powers (not necessarily distinct). The numbers $d_i$ and $e_j$ are uniquely determined by $A$ and $B$ respectively. So by the given condition, we must have $r = s, d_i = e_i, \forall i.$ This shows that $A$ and $B$ are isomorphic.

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    $\begingroup$ "So by the given condition..." That's a big jump there. $\endgroup$ – Thomas Andrews Dec 29 '14 at 15:00
  • $\begingroup$ @ThomasAndrews: what I wrote was wrong. we need prime powers. other wise it need not be true,e.g. $d_1 = 6, d_2=5, e_1=2, e_2 = 3, e_3 = 5.$ now I've fixed it. Thanks for pointing out. $\endgroup$ – Krish Dec 29 '14 at 15:34

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