Let $G$ be a finite group of even order. Then which of the following statements is correct? 
Let $G$ be a finite group of even order. Then which of the following
  statements is correct ?
$1.$ The number of elements of order $2$ in $G$ is even
$2.$ The number of elements of order $2$ in $G$ is odd
$3.$ G has no subgroup of order $2$
$4.$ None of the above.


My attempt : I take $G=  K_4   $, then option $1$ is correct
Is it true  ?
Any hints/solution
Thank you!
 A: Option 1 is false, actually the group $K_4$ has $3$ elements of order $2$ 

The group $G$ must has an element of order $2$ is true, and it is an easy exercise to you!
If $a \neq a^{-1}$, then $\vert a \vert$ is not $2$, so this type of elements comes in pairs, which make an even count. Call this count as $k$, then
$$\vert G \vert =\underbrace{\vert \{e\} \vert}_{1} + \underbrace{\vert \{a \in G :a \neq a^{-1} \} \vert}_{k\;\; \text{an even count}} +\text{number of remaining order $2$ elements}$$ 
But Whole count is even, so the last count must be ODD!
Option $3$ is obviously false!
A: As given that group $G$ is of even order so number of elements of order $2$ should be odd since any element of order $2$ is self inverse and any other element of order more than $2$ must appear in the group with its inverse element also so total number of elements in finite group of even order is number of self inverse elements+number of nonself inverse elements clearly number of non self order elements are even and in any group identity element is always self inverse so number of elements of order $2$ should be odd(since self inverse elements other than identity are of order $2$ only). Therefore option $1$ is false, for example you can consider $S_3$ group of symmetry on $3$ symbols it has $3$ elements of order $2$. So option $2$ s correct. Option $3$ can be easily discarded using the Syllow theorem.
