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First, suppose $G_1, G_2$ are two finite groups and $|G_1|=|G_2|$.

If for each $n\in\mathbb N_+$, we have $\#\{g_1\in G_1,o(g_1)=n\}=\#\{g_2\in G_1,o(g_2)=n\}$.

My question:

$1$. Can we always have $G_1\cong G_2$?

$2$. What if $G_1$ and $G_2$ are infinite groups?

$3$. What can be used to characterize a group?

For example, multiplication table, generator and relation, its all subgroups(especially Sylow p-sungroups), etc while character table can't.

Thanks for your time and effort.

Andrews
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    For your second question, if $G_1$ and $G_2$ are both infinite torsion-free groups, then the number of elements of any given order is either 1 (for order 0) or 0 for any order in $\mathbb{N}$. But of course one can pick any two different torsion-free non-isomorphic groups for this. – Carl-Fredrik Nyberg Brodda Nov 06 '19 at 08:52
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    There are not only two but three groups of order 16 where the number of elements of order $n$ is the same in all three groups for all $n$. See https://math.stackexchange.com/questions/62331/three-finite-groups-with-the-same-numbers-of-elements-of-each-order – MJD Nov 06 '19 at 14:45

1 Answers1

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  1. No, this cannot be concluded, see this MO answer.
  2. There it's easy to see that it doesn't work: Take $G_1 = \mathbb Z$ and $G_2 = \mathbb Z^2$.
  3. This is a hard question. The classification of finite simple groups is one of the greatest accomplishments of modern mathematics and took thousands of pages of academic writing to establish. There really is no satisfying answer to this question.

All in all, the theme is that group theory is hard. There are many groups, and most of them are crazy. Checking that two given groups are isomorphic is undecidable.

Levi
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