Why a group that has four elements of order two does not exist? My professor said, if you remove identity element and self inverses, the number of elements of order two should be odd. So, in a group, the number of elements of order two can't be four.Why so?
 A: Preliminary remarks (motivated by bof's comment below). It is difficult to parse what you say your professor says for a few reasons. First, the identity of a group is its own inverse, so "identity and self-inverses" is redundant. Second, an element of a group is a self-inverse if and only if it is the identity or has order two. So if you "remove self-inverses" then there are no elements of order two left.
In any case, here are the facts:
Fact 1. If $G$ is a finite group of odd order, then $G$ has zero elements of order 2.
Fact 2. If $G$ is a finite group of even order, then $G$ has an odd number of elements of order 2.
Fact 3. If $G$ is an arbitrary group with a finite, but nonzero, number of elements of order 2, then $G$ has an odd number of elements of order 2.
So if we put it all together, we get the following description.

If $G$ is a group, then exactly one of the following holds.

*

*$G$ has no elements of order 2.

*$G$ has infinitely many elements of order 2.

*$G$ has an odd number of elements of order 2.


Note that Fact 3 generalizes Fact 2, if you assume the $p=2$ case of Cauchy's Theorem, which says that a finite group of even order has an element of order 2. However, the $p=2$ case of Cauchy's Theorem follows directly from Fact 2. So this justifies giving separate proofs of Facts 2 and 3.
So let's start the proofs.

Proof of Fact 1. This follows from Lagrange's Theorem, which implies that the order of an element in a finite group always divides the order of the group.


Proof of Fact 2. Partition $G$ into three pieces:
Piece 1: the identity element
Piece 2: the elements of order greater than 2
Piece 3: the elements of order 2
There are an even number of elements in piece 2 since every element in piece 2 can be paired with its inverse, which is also in piece 2 and is not equal to the original element. (Here we use the fact that $x=x^{-1}$ iff $x$ has order at most 2.)
So the total number of elements in pieces 1 and 2 is odd. Since $G$ has even order, the number of elements in piece 3 is also odd.


Proof of Fact 3. (See this question: the number of elements of order 2 in an infinite group. I will repeat the argument by Mikko Korhonen.)
Let $G$ be a group and let $X$ be the elements of order at most 2. Assume $G$ has an element $t$ of order 2 (so $t\in X$). Partition $X$ into two pieces. Piece 1 is the elements in $X$ that commute with $t$, and piece 2 is the rest. Then we can pair each $x$ in piece 1 with $xt$, and we can pair each $x$ in piece 2 with $txt^{-1}$. (One must check that this is a well-defined pairing, i.e., if $x$ is in piece 1 then $xt$ is in piece 1 and distinct from $x$; and if $x$ is in piece 2 then $txt^{-1}$ is in piece 2 and distinct from $x$.) So both pieces have an even number of elements, hence $X$ has an even number of elements. Removing the identity, we get an odd number of elements of order 2.

