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Are the numbers of elements of two distinct prime orders not equal in finite groups?

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2 Answers 2

The answer is "no":

One can construct a counterexample of the form $G\times H$. Say $p$ and $q$ are distinct primes, and for some integer $m$ one can find two groups $G, H$ with with $q\nmid|G|$ and $m$ elements of order $p$ in $G$, and $p\nmid|H|$ and $m$ elements of order $q$ in $H$. Then $G\times H$ has $m$ elements of order $p$, and $m$ of order $q$.

Now take $p=5$ and $q=3$. Then $G=C_{11}\rtimes C_5$ has $44$ elements of order $5$ (every element not in $C_{11}$), and $H=C_{21}\rtimes C_3$ has $44$ elements of order $3$ ($2$ in $C_{21}$ plus all $42$ outside it). So $G\times H$ has 44 elements of order $3$ and of order $5$.

To check this in Magma:

P:=DirectProduct(SmallGroup(55,1),SmallGroup(63,3));
assert #[g: g in P | Order(g) eq 5] eq 44;
assert #[g: g in P | Order(g) eq 3] eq 44;
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This is very clever. –  Todd Trimble Dec 18 '10 at 12:46

Consider $C_6$. It has 1 element of order 1, 1 element of order 2, 2 elements of order 3, and 2 elements of order 6.

Notice here that the numbers of elements of order 2 and of order 3 differ.

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I interpret the question as wanting a finite group G and distinct primes p and q where the number of elements of order p and of order q in G are the same and nonzero. Gerhard "Ask Me About System Design" Paseman, 2010.12.17 –  Gerhard Paseman Dec 18 '10 at 7:22
3  
Or a proof that there is no such finite group. Gerhard "Then The Other Shoe Drops" Paseman, 2010.12.17 –  Gerhard Paseman Dec 18 '10 at 7:24

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