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If $G$ is a group of even order, then show that there exists an element $a≠e$, $e$, the identity in $G$,

such that $a^2 = e$.
asked in Set Theory & Algebra by Veteran (59.4k points)
edited by | 571 views
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Just one additional information order of an element divides the order of Group.
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This might help ...

2 Answers

+24 votes
Best answer

I will proove it using contradiction. Assuming no element has order 2. i.e. $a^2 \neq e$ for any non-identity element a, means 

$a \neq a^{-1}$ for any (non-identity element) a.

Further rewriting meaning: the inverse of any element is not that element itself, it is someone else.

But i want to somehow show that at least one element has inverse as its own. I am trying a method lets see if it works :).

 i will select each element from the set and will check inverse of each element, and in this process, as soon as i encounter any element having  $a^2 \neq e$ then i am done.

My Goal: to show G has atleast 1 element as its own inverse.

Let $|G| = 2n$. Then we take out the identity and have  2n-1 elements to choose from.

Step1: select an element, if it is its own inverse then i am done.

Step2 (otherwise) : if inverse of a is not a and its b. then throw a and b out. (bcoz if 'a' inverse is 'b' then 'b' inverse also 'a', and inverse of an element is unique.)                  

(Notice we always throw one pair) 

In worst case i will ended up throwing each element but one element will left as it is, bcoz total number of elements are odd (2n-1) and we always throw a pair of two (even).

Now question is, What is inverse of that element ? 

it has to be its own inverse, it can not map to inverse of any other element bcoz inverse is unique. And moreover it can not be inverse of identity element bcoz inverse of identity is identity itself.

Finally, i can say there exist one non-identity element 'a' of order 2.

Yes, it worked !

Hence Proved !

answered by Boss (15.5k points)
selected by
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Nice approach to solve such problems.Thanks :-)

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is it possible here |G|=2n+1 elements , then after removing identity element we will have 2n elements ?
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they given group has even order ...
+7 votes

Now what i can say for group of even order has even no. of elemnt in it at-east one time identity elemnet come to diagonal place exept for identity element itself, at that time a2 = e is happen .

Point -1: Identity element has its own inverse but we cannot take it.[ Given in question  a≠e]

Point-2 : since even numbers total but we left with odd numbers (2n-1). an we know if a is inverse of b then b is also inverse of a.

So 2n-2 elements are got pairing with thier inverse so 1 left element is its own inverse.

In image 4 is that element.

answered by Veteran (59.7k points)
edited by


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