GATE CSE 1992 | Question: 14a

Question

If $G$ is a group of even order, then show that there exists an element $a≠e$, the identity in $G$, such that $a^2 = e$.

Comments

Just one additional information order of an element divides the order of Group.

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It not only divides but it is equal

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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.

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$\color{red}{\text{Detailed Video Solution:}}$ https://youtu.be/Ewdz0FQJ14w?feature=shared

Answer 1

Detailed Video Solution:

Detailed Video Solution: GATE CSE 1992 - Group of Even Order

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I will prove 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 an inverse as its own. I am trying a method; let's see if it works :).

I will select each element from the set and check the inverse of each element. 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 the 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 the inverse of an element is unique$)$                  

(Notice we always throw one pair) 

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

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

It has to be its own inverse. It cannot map to the inverse of any other element because the inverse is unique. Moreover, it cannot be the inverse of the identity element because the inverse of identity is identity itself.

Finally, I can say there exists one non-identity element $a$ of order $2$.

Yes, it worked!

Hence Proved!

Please watch this video to solve this question efficiently and clearly.

 

Detailed Video Solution: GATE CSE 1992 - Group of Even Order

 

Comments

This explanation is amazing..Thanku @Sachin Mittal 1 Sir.

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very nice explaination sir

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I would give this answer $+100$ up votes if I could.

Answer 2

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.

Answer 3

Think of this like the theorem in Graph Theory that the number of vertices with odd degree must be even.

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To prove: If G is a group of even order, then there exists an element a, such that order of a is 2. And, a can't be the identity.

 

We know that if an element is its own inverse, its order is 2.

ie, if $a=a^{-1}$ then $a^2=e$

 

Say the group has 2n elements, (hence always even order).

The inverse of the first element is the last element.

The inverse of the second element is the second last element.

The inverse of the third element is the third last element.

So on...

We know, we can make pairs out of even numbers easily. For 2n numbers, we can have n pairs.

BUT

Somewhere in this sequence, we'll encounter the identity element. The inverse of the identity element is the identity element itself!

That means all the pairs aside, we have a standalone element a, and an identity element.

 

Since every element must have an inverse in a group, and no two elements have the same inverse => a must be it's own inverse.

So, order of a is 2.

 

Proved.

Answer 4

Given: |G| = even.

Approach: here I am using non-isomorphic group structure concept to prove that if the given group is of even order then there exists an element ‘a’ such that  a^2 = e or a^-1 = e

here a^-1 means inverse of element ‘a’
and ‘e’ means identity element

let’s take a group of |G| = 4 = even order:
then there are only two non-isomorphic group structures of order 4 possible:

1^st structure template:- G = ({e, x, y, z}, *) where, x^-1 = x, y^-1 = y, z^-1 = z and obviously e^-1 = e

note that here inverse of ‘x’ is ‘x’ this means x^2 = e. And similarly:
y^2 = e
z^2 = e
so, in this structure (group of order 4) other than ‘e’ we have three more elements such that a^2 = e

2^nd structure template):- G = ({e, x, y, z}, *) where x^-1 = y, y^-1 = x, z^-1 = z and e-1 = e

note that here also the inverse of ‘z’ is ‘z’ i.e; z^2 = e
so, this structure (group of order 4) also has an element ‘z’ other than ‘e’ such that a^2 = e

now let’s take |G| = 2 = even order:
here only one non-isomorphic group structure of order 2 is possible: 

structure template:- G = ({e, x}, *) here x^-1 = e and of course e^-1 = e
so, here also apart from identity element ‘e’ there exists an element ‘x’ such that x^2 = e

so, similarly for any other even order group, there will exist at least one element other than identity element ‘e’ such that a^2 = e.

https://math.stackexchange.com/questions/961286/non-isomorphic-groups