# Can every element of a group be written as the product of two non-identity elements of the group?

By part of the definition,

two elements in a group can be put together with the group operation to obtain a third element that is also an element of the group.

However, I am wondering if the converse is also true. So the new statement would be:

For every element in the group, it can be written as the result of two non-identity elements of the group using the group operation.

So here we are not considering the element itself with the identity. Is there any counterexample? Thanks.

• This is true for every non-trivial group, but is not true for the trivial group $G = \{0\}$. Apr 17, 2020 at 2:45
• @Omnomnomnom $C_2$ is also a counter-example, as pointed out by Jacob (with a proof that these two are the only counterexamples). Apr 17, 2020 at 22:49
• @verret somehow I had thought that only on of the elements had to be distinct from the identity, thanks for the correction Apr 17, 2020 at 23:13

Yes, there are counterexamples. Apart from the trivial group, there is also the two element group $$C_2=\{e,a\}$$, where $$a^2=e$$. In this group, $$a$$ is the unique non-identity element and $$a^2=e$$ so $$a$$ cannot be written as a product of non-identity elements.
These are the only counter-examples. Indeed, let $$G$$ be group of cardinality at least three and let $$g\in G$$. We write $$g$$ as a product of two non-identity elements. If $$g=e$$, then take $$h\neq e$$ and we have $$g=e=h*h^{-1}$$. If $$g\neq e$$, then there exists $$h\in G\setminus\{e,g\}$$ and $$g=h*(h^{-1}g)$$, with neither $$h$$ nor $$h^{-1}g$$ being the identity.
• @DerekHolt As Jacob said, it is the non-identity element in $C_2$ which cannot be written as a product of two non-identity elements, even allowing for repeated elements. Apr 17, 2020 at 22:42
Your claim is true as soon as $$|G|\ge 3$$, as a corollary of this general result by taking $$H=\{e\}$$.