I have examined some of the tables of various groups of the form $\mathbb Z_n$ and I noticed that $\mathbb Z_3$ has a subgroup of order 2 which doesn't divide 3. that subgroup is $\{1,2\}$ under multiplication.

now I checked wikipedia and they stated a proof there of lagrange's theorem. although this invalidity isn't relevant to $\mathbb Z_3$, it stated that $\forall a\operatorname{f}(x)=ax$ is bijective. this isn't generally true because in some groups there are zero-divisors(this term is related to ring theory but the meaning is hopefully clear enough if you replace zero by an absorbing element).

now it could be that the theorem relies on there being an inverse element for every element but zero doesn't have one so the theorem doesn't work on $\mathbb Z_3$ for that reason. also the theorem does works on $\mathbb Z_3 / \{0\}$.

now I just want to know is this a contradiction, or that the groups the theorem talk about are groups with no absorbing element.

  • 3
    $\begingroup$ You might need to check the definition of "subgroup". $\endgroup$ Jul 7, 2023 at 14:12
  • 2
    $\begingroup$ The theorem relies on the two group operations being the same. $\endgroup$ Jul 7, 2023 at 14:19
  • $\begingroup$ The only group that has an absorbing element is the trivial group. By definition, an absorbing element $z\in G$ has to satisfy $g z = z $ for all $g\in G$, and since in a group every element is invertible, we get $g=1$ for all $g\in G$. $\endgroup$ Jul 7, 2023 at 17:53

1 Answer 1


$(\mathbb Z_3,\times)$ is not a group, so the theorem does not apply. $(\mathbb Z_3 \backslash \{0\},\times)$ is indeed a group, but it cannot be a subgroup of $(\mathbb Z_3,\times)$ because, again, the latter is not a group.

$(\mathbb Z_3,+)$ is a group (and the theorem applies without any issue), but $\mathbb Z_3 \backslash \{ 0\}$ is not stable by $+$, so it is not a subgroup.

A group $(G,*)$ does not have "zero divisors" for the law $*$, because every element is invertible.


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