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Let $G$ be a group with the property $G=G_e\dot{\cup} G_o$ with $G_oG_o\subseteq G_e\leq G$.

($\dot{\cup}$ denotes disjoint union, $\leq$ is subgroup notation, and $G_o^{-1}=\{x^{-1}: x\in G_o\}$)

We observe that

(1) $G_o^{-1}=G_o$, $G_oG_o=G_e$, $G_oG_e=G_eG_o=G_o$;

(2) $|G:G_e|=2$, $|G|=2|G_o|=2|G_e|$;

(3) $(\mathbb{Z},+)$, $S_n$ and $\mathbb{Z}_{2m}$ enjoy the property.

Now, is there any characterization for such groups?

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  • $\begingroup$ You haven't told us what $G_e$ stands for, nor $G_O$, nor $G_o$. $\endgroup$
    – Gerry Myerson
    Commented Jan 31, 2016 at 11:25
  • $\begingroup$ Also you have used the tag finite-groups, but not said whether you are assuming that $G$ is finite. $\endgroup$
    – Derek Holt
    Commented Jan 31, 2016 at 11:36
  • $\begingroup$ Because we have both finite and infinite classes of such groups (Derek). Why is it not true (Geoff)? $\endgroup$
    – M.H.Hooshmand
    Commented Jan 31, 2016 at 19:22
  • $\begingroup$ @M.H.Hooshmand I think the subject which you mentioned is important, since it is related to graded structure for group rings.Please see the paragraph after example 2.2 of this note(and its consequences, in particular proposition 2.4): hindawi.com/journals/aaa/2012/729745 $\endgroup$
    – Ali Taghavi
    Commented Mar 1, 2016 at 10:53

1 Answer 1

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Group $G$ has subsets $G_e$ and $G_o$ as mentioned above, iff $G$ has a normal subgroup of index 2. If $G_e$ is a normal subgroup of index 2, then $G_e$ and $G_o=xG_e$, $x\notin G_e$, satisfied the condition of the problem.

Now, suppose such subsets $G_e$ and $G_o$ exist. Since $G_e$ is a subgroup, and $G_e\cap G_o=\emptyset$ $G_o^{-1}\subseteq G_o$, and so, $G_o^{-1}=G_o$. Therefore $G_oG_o=G_e$. This implies that $G_ex=xG_e=G_o$ for $x$ not in $G_e$.This leads $G_e$ to be a normal subgroup of index 2.

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