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I know that if two groups are isomorphic, they have same number of elements of order n. But will the converse holds? I think it won't as there are many group characteristicss like abelian or cyclic will also matter. But I can't come up with example.

ogirkar
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  • Take one of those characteristics, like abelian. Can you come up with two groups of the same size where one is abelian and the other isn't? What small groups do you know of? – Chessanator Jun 01 '19 at 21:19
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    @Chessanator The OP wrote about the orders of the elements, not about the orders of the groups. – José Carlos Santos Jun 01 '19 at 21:21

1 Answers1

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Not in general. The group $\mathbf{C}_3\times\mathbf{C}_3\times\mathbf{C}_3$, where $\mathbf{C}_3$ is the cyclic group of order $3$, has $27$ elements, of which one has order $1$ and 26 have order $3$.

Now consider the group of all $3\times 3$ matrices of the form $$\left(\begin{array}{ccc} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1 \end{array}\right)$$ with $a,b,c$ taken in $\mathbb{Z}/3\mathbb{Z}$ (this is a “Heisenberg group”). It is not hard to verify that $$\left(\begin{array}{ccc} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1 \end{array}\right)\left(\begin{array}{ccc} 1 & x & z\\ 0 & 1 & y\\ 0 & 0 & 1 \end{array}\right) = \left(\begin{array}{ccc} 1 & a+x & c+z+ay\\ 0 & 1 & b+y\\ 0 & 0 & 1 \end{array}\right)$$ and therefore that every element has order either $1$ or $3$. So it has exactly one element of order $1$, and 26 of order $3$.

But this group is not abelian, so it cannot be isomorphic to $\mathbf{C}_3\times\mathbf{C}_3\times\mathbf{C}_3$.

In fact, there are examples of order $p^n$ for any odd prime $p$ and $n\geq 3$, and examples of order $2^n$ for any $n\geq 4$. With these, you can construct examples for groups of any order that is divisible by the cube of an odd prime or by $16$.

Arturo Magidin
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