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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 $a^2 = e$.
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Just one additional information order of an element divides the order of Group.
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This might help ...

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It not only divides but it is equal
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nice video puja mam
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If order of a group is even, there will be an element which is inverse of itself.

=> a * Inverse = e(Identity)

=> a * a = e ('a' is such element)

=> a^2 = e.

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) $a.$

Rewriting the statement: the inverse of any element is not that element itself, it is something 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 at least $1$ element as its own inverse.

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

Step 1: Select an element, if it is its own inverse then I am done.

Step 2 (otherwise): If inverse of $a$ is not $a$ and is $b,$  throw $a$ and $b$ out. $(\because$ if $a$ inverse is $b$ then $b$ inverse is $a$, and inverse of an element is unique$)$

(Notice we always throw one pair)

In worst case I will end up throwing all but one element, because total number of elements is odd $(2n-1)$ and we always throw a pair of two (even).

Now question is, what is the inverse of that element?

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

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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 ...
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Let's say we have a group G of even order

So it has even number of elements.

Now the order of each element of a group have to be divisor of the order of the group.

So, let us assume we have an element $a$ with order 2k and this means $a^{2k}=e$

Now I can rewrite the above as $(a^k)^2=e$

Now since a group has closure property, $a^k$ would surely belong to one of the element of the group, say it is x.

Now $x^2=e$ and we have proved what we wanted.
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I have a query here.

For a finite group(G), is it necessary that every prime divisors of order G(|G|) , will be the order of one/more elements of G?
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@you mean to say that consider a group of order 15. Prime divisors of 15 are 3,5. So the order of one or more elements of this group G would be one of 3 or 5?
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Yeap..That's the query whether 3 and  5 would be order of one or more elements?
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we can prove like this also -

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.

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