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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.

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    $\begingroup$ This is true for every non-trivial group, but is not true for the trivial group $G = \{0\}$. $\endgroup$ Apr 17 '20 at 2:45
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    $\begingroup$ @Omnomnomnom $C_2$ is also a counter-example, as pointed out by Jacob (with a proof that these two are the only counterexamples). $\endgroup$
    – verret
    Apr 17 '20 at 22:49
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    $\begingroup$ @verret somehow I had thought that only on of the elements had to be distinct from the identity, thanks for the correction $\endgroup$ Apr 17 '20 at 23:13
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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.

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    $\begingroup$ While the answer is mostly fine I find the "yes" at the start highly confusing. Why "yes" when you then proceed to give a counterexample to what is asked for (granted in a corner case but still). I'd just remove it. Or, say "yes, for all groups of cardinality at least three, but not for those of cardinality one or two." or something like this. $\endgroup$
    – quid
    Apr 17 '20 at 11:43
  • $\begingroup$ @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. $\endgroup$
    – verret
    Apr 17 '20 at 22:42
  • $\begingroup$ @verret Yes of course! $\endgroup$
    – Derek Holt
    Apr 18 '20 at 7:27
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Your claim is true as soon as $|G|\ge 3$, as a corollary of this general result by taking $H=\{e\}$.

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