If an abelian group has more than 3 elements of order 2 then it must have at least 7 elements with order 2. 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.         
 A: 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.
A: Suppose we have two elements of order two $\{a,b\}$. Then $ab$ is also of order two, since $ab\cdot ab = ab\cdot ba = a\cdot b^2 \cdot a = a^2 = e$
Similarly if we have four elements of order two, then there are two options: either one of the elements is the product of two of the others, or not. In the first case, elements $\{a,b, ab, c\}$ we can see that $\{ac,bc, abc\}$ are also all of order two, giving $7$ elements. In the independent variable case, there may be even more elements of order two, or it may be a variation of the $7$ element case eg $\{a,b, ac, bc\}$ - note that $ac\cdot a = c$ so we are back to the same case. 
Essentially cross-multiplying different order-2 elements will always produce order-2 elements in such groups, giving the "magic numbers" for the count of such elements.
A: The elements of order $2$ together with the identity element in an abelian group form a subgroup. That subgroup is a vector space over the field $\mathbb{F}_2$ with two elements (under the action induced by the usual action that makes any abelian group into a $\mathbb{Z}$-module). A finite vector space over $\mathbb{F}_2$ has finite dimension $d$, say, and hence contains $2^d$ elements. All but one of these elements have order $2$. So the number of elements of order $2$ in an abelian group is either infinite or it is $2^d-1$ for some $d \in \mathbb{N}$.
A: Assume you have 4 distinct elements of order 2. $x,y,z,w$.   Then $xy\ne xz \ne xw$, and each of those have order 2, so there's your 7. 
A: For finitely generated groups it seems to follow directly from the structure theorem.
$G$ is isomorphic to a product of cyclic groups of prime power and $\mathbb Z$, the elements of order $2$ all have $0$ in the coordinates corresponding to all the groups in the product that are not of the form $\mathbb Z_{2^n}$.
On the other hand every group $\mathbb Z_{2^n}$ has exactly one element of order $2$. Hence if $G$ has $k$ factors of the form $\mathbb Z_{2^{i}}$ then it has $2^k-1$ elements of order $2$ (because the coordinates corresponding to each such factor has to be the element of order $2$ or the indentity and it is not possible for all coordinates to correspond to the identity)
