Let G be an abelian group of order 2n, where n is odd. Prove that G has exactly one element of order 2. Does the same conclusion hold if G is not necessarily commutative?
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Do you know the structure theorem for finite abelian groups? – Seth Apr 29 '14 at 21:48
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For the nonabelian groups, there exists an element of order 2, but uniqueness not guaranteed. – Sungjin Kim Apr 29 '14 at 21:51
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You can show that since $G$ has even number of elements, there is an odd number of elements of order 2, and in particular there is one element of order 2. Show yourself that's true (Hint: for element $x$ of order higher than two, $x \neq x^{-1}$).
Lets assume there are $k$ number of elements of order 2, $k>1$. So lets take all elements of order 2: $x_1, x_2, x_3, \cdots , x_k$, $k$ being odd. Lets look in the multiplication of all of them: $x_1 \cdot x_2 \cdots x_k$.
So since the order of $x_i=x_i^{-1}=2$, we know that for every element in the multiplication, its inverse is also there. So in some rearrangement we can do (since $G$ is abelian), we can rearrange the it so: $x_1 \cdot x_1^{-1} \cdot x_2 \cdot x_2^{-1} \cdots x_k$. Notice though that we arrange it in pairs: element multiplied with its inverse, but we have $k$ - odd elements. So we get the result of this multiplication is some $e\cdot e \cdot \cdots x_m=x_m$. So we get an element $x_m$, which isn't of order 2, otherwise we could pair it with its inverse. Thus contradiction.
P.S. I just attempted this proof. Not sure if it's correct - would like feedback. Also, I'm not familiar with the abelian structure theorem.
Sorry if my attempt was in futile.
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This answer is wrong. The correct way is to use Lagrange's Theorem. – YuiTo Cheng Jun 27 '22 at 05:35
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Hint: Use the structure theorem for finite abelian groups. For the converse, consider the permutation group $S_3$ which has order $6$ and has 3 elements of order $2$.
user2566092
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