ISI2018-MMA-15

Question

Let $G$ be a finite group of even order. Then which of the following statements is correct?

• A. The number of elements of order $2$ in $G$ is even
• B. The number of elements of order $2$ in $G$ is odd
• C. $G$ has no subgroup of order $2$
• D.  None of the above.

Answer 1

Answer is B.

Since the group is of even order and the identity is the inverse of itself, therefore, there are odd number of elements (other than identity).

Also, if there is some element which is not inverse of itself, then we can pair such elements with their inverses. This will leave us with odd number of elements which have to be their own inverses.

Comments

here why did you not include identity

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unable to understand example?

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what about the elements with order greater than 2? They have not been included.

Answer 2

By sylow’s third theorem, if |G| / p^k    where “p” is the prime number of any power k then,

Number of subgroups of order p^k will be 1+kp.    Here prime number mentioned is 2 thus 1+2p ie odd subgroups

Hence option B is correct

Answer 3

Number of elements = even = 2n

according to the property og group we have unique identity

let it be ‘e’

so out of 2n elements one is ‘e’

remaining elements = 2n -1

For a element to be of order 2 it should be a*a = e

hence a = a^-1

so the maximum these type of elements we can have is 2n-1 which is odd.