# Let $G$ be a finite abelian group with elements $a_1,a_2,\dots,a_n$. If $G$ has more than one element of order $2$ then $a_1a_2\dots a_n=1$.

Let $G$ be a finite abelian group with elements $a_1,a_2,\dots,a_n$.
If $G$ has more than one element of order $2$ then $a_1a_2\dots a_n=1$.

Attempt
Clearly, if $a_i$ is not of order 2, the inverse of $a_i$ must be in the product.
So the elements left are all of order $2$ or identity, say $b_1,b_2,\dots,b_m$
Since $b_i^2=1$ for $i=1,\dots,m$.
Let $H=\{1,b_1,\dots,b_m\}$
Then $H \cong C_2\times C_2 \times\dots \times C_2$.

For $C_2\times C_2$, it is indeed $V$-group $\{1,a,b,ab\}$. Clearly, $1abab=1$.
Assume the result holds for direct product of less than $k$ cyclic groups of order $2$. Let $H$ be a direct product of $k$ cyclic groups of order $2$.
Write $H=\langle b_1\rangle \times \dots \times \langle b_k\rangle$.
Consider $K=\langle b_1\rangle \times \dots\times \langle b_{k-1} \rangle$.
Then the result holds for $K$. Now I need to relate this result to $H$.

• The solution is fine except for the last sentence. I do not see what your argument is there. – Tobias Kildetoft Nov 11 '16 at 10:46
• @TobiasKildetoft Ouh I realized my mistake. Then I think my solution is not complete – Alan Wang Nov 11 '16 at 10:47
• I agree it is not complete, but you are on the right track by writing the relevant elements as a direct product. I would consider each entry in such a tuple separately in the product and do a bit of counting. – Tobias Kildetoft Nov 11 '16 at 10:51
• @TobiasKildetoft I continue my attempt by using induction but have no idea how to relate the induction hypothesis to complete the induction. – Alan Wang Nov 11 '16 at 11:00

To finish your inductive proof, first look at the elements in $H$ that doesn't contain a factor $b_k$ (we'll include $1$ here for counting purposes). This is just the product of all elements of $K$, so by the inductive hypothesis, they all multiply to $1$.

Now look at the elements of $H$ that do contain a factor $b_k$. This is the same as above, except that each element now carries an additional factor $b_k$ (remember to include the lone $b_k$ as well). The product of all those elements is therefore $b_k^{|K|}$ times the product of all elements of $K$, which simplifies to just $b_k^{|K|}$ by the inductive hypothesis.

Lastly, note how many elements there are in $K$, and you see that $b_k^{|K|} = 1$.

One could also try to appeal to some sort of symmetry moral with this problem (this is not a valid proof, by any measure, but I personally like to think along these lines when pondering the eternal question "yeah... but why?"). Note that $(a_1a_2\cdots a_n)^2 = 1$, so $a_1a_2\cdots a_n$ is either the identity or some degree-2 element. If there is only one degree $2$ element in the group, then there's nothing wrong with $a_1a_2\cdots a_n$ being that one element. But if there are more than one, how would the group know which one to pick? The group being abelian means that there is no algebraic property that distinguishes any of the order $2$ elements, but "being the product of all the elements in the group" is a pretty distinguishing feature. The most (/ only?) consistent choice for $a_1a_2\cdots a_n$ is therefore $1$.

• Is it $|K|=2^{k-1}$, so $b_k^{2^{k-1}}=1$? – Alan Wang Nov 11 '16 at 11:27
• A bit confusion here. Isn't it $|K|=|b_1|\dots |b_{k-1}|=2^{k-1}$? – Alan Wang Nov 11 '16 at 11:29
• @AlanWang Exactly (I misread your exponent with my previous comment). So the product of all elements in $H$ that doesn't have a $b_k$ in them is $1$, and the product of all elements in $H$ that does have $b_k$ in them is $b_k^{2^{k-1}} = 1$. Therefore the product of all the elements in $H$ is also $1$. – Arthur Nov 11 '16 at 11:30

Consider $H=\{g\in G:g^2=1\}$; then $H$ is a subgroup of $G$. Moreover, the product of the elements not in $H$ is $1$, because each element has there its inverse (distinct from it).

Thus you can assume, without loss of generality, that $H=G$. Therefore $H\cong C_2^n$. If $n>1$, there is an automorphism that swaps any two components. Since the product of all elements is invariant under automorphisms…