This is an exercise in Gallian, Contemporary Abstract Algebra. Intuitively, I cannot believe the statement is true. It is essentially saying there is no abelian group with 4, 5 or 6 elements of order 2 regardless of how large the group is. I have tried to suppose a group had exactly 4 (or 5 or 6) elements of order 2 in hopes of reaching a contradiction. I know that if a group is cyclic then it has exactly 1 element of order 2. I know that an element of order 2 is its own inverse and it is not the identity. I have observed that indeed in $C_2 \times C_2$ there are 3 elements of order 2 and in $C_2 \times C_2 \times C_2$ there are 7 elements of order 2. However, none of this is convincing me that the statement is true.
[Math] If an abelian group has more than 3 elements of order 2 then it must have at least 7 elements with order 2.
group-theory
Best Answer
Alan's post has been deleted, sadly, because it was a concrete answer and very nearly there. This is to encourage a response to the comment which caused its deletion.
Adam noted that if there were four elements of order $2$ we could call them $x,y,z,w$ and postulated $xy\neq xz\neq xw$ to give three extra elements of order $2$.
A comment pointed out that we could have $xy=z$, so these need not be distinct. If this is so, then we also have $xz=y, yz=x$ and the proof can be completed with $xw, yw, zw$. These are distinct from each other, and note that in this case $xw=y\implies xw=xz$ etc to show that these are distinct.
This is Adam's proof, only filled out by me, so please, if you have the rep, vote to undelete his answer, so he can amend the answer and get the credit for a good observation.