Let $G$ be a group and $H$ be a subgroup of $G$. The left $H$-cosets are the sets $gH, g \in G$. The set of left $H$-cosets is the quotient set $G/H$. The right $H$-cosets are the sets $Hg, g\in G$. The set of right $H$-cosets is the quotient $H \setminus G.$ The set of $G$ is the union of left(respectively, right) $H$-cosets, each of which has $\left|H\right|$ elements. We deduce that the order of $H$ divides the order of $G$, and that the number of left $H$-cosets equals the number of right $H$-cosets.

What I wrote can be find in Groups and symmetries - Yvette Kosmann-Schwarzbach.

What should I understand from that statement ?

is ok if I say that : $$G=gH \cup Hg?$$

$$\left|G/H\right|=\left|H \setminus G\right|?$$ What means that the number of left $H$-cosets equals the number of right $H$-cosets?

And why we deduce that the order of $H$ divides the order of $G$?

Thanks :)

  • $\begingroup$ The quoted "argument" does not work. For more details, consult a better book on group theory, or look at math.stackexchange.com/questions/344176 $\endgroup$ Commented Mar 30, 2013 at 20:29
  • $\begingroup$ I have never seen the notation $H\setminus G$ to denote right cosets. It seems to collide with the set theoretical notation of relative complementation $H\setminus G$. Maybe you can specify the coset space in words, or symbolically, as $G/H_R$ or $G/H_L$. Of course, if $H$ is normal there is no ambiguity in $G/H$. $\endgroup$
    – Pedro
    Commented Mar 30, 2013 at 20:31
  • $\begingroup$ That stupid notation is mentioned exactly once that I can recall from Dummit & Foote. Thankfully it is (deservedly) uncommon. $\endgroup$
    – Tyler
    Commented Mar 30, 2013 at 20:39
  • 9
    $\begingroup$ @TylerBailey: Dear Tyler, This notation is not uncommon at all in certain areas of mathematics, especially parts of representation theory, geometry, and the theory of automorphic forms, where one often has to consider group actions on homogeneous spaces, or (what is closely related) double coset spaces, where the issue of left vs. right cosets and left vs. right actions starts to matter. E.g. there are many mathematicians for whom the notation $\Gamma \backslash G / K$ (with that particular notation, and pronounced "$G$ mod $K$ mod $\Gamma$) has immediate and enormous resonance. Regards, $\endgroup$
    – Matt E
    Commented Mar 30, 2013 at 20:41
  • 2
    $\begingroup$ @MartinBrandenburg: Dear Martin, The argument seems correct if $G$ is finite, which is likely the main focus of the text. Regards, $\endgroup$
    – Matt E
    Commented Mar 30, 2013 at 20:44

3 Answers 3


You can produce a bijection of the left coset space $L_G=\{gH:g\in G\}$ onto the right coset space $L_R=\{Hg:g\in H\}$ by $gH\mapsto Hg^{-1}$. Regarding the fact that $|H|\;\mid \;|G|$, use the fact that each coset has the same cardinality: Take $zH$ and $xH$ different cosets. Then $zH\to xH:u\mapsto xz^{-1}u$ is a bijection. In particular, $|H|=|gH|$ for any coset, so $|G|=m|H|$ for $m$ the number of cosets in the coset space, usually denoted by $m=|G:H|$. It is also crucial that given two cosets $zH,xH$, $zH\cap xH\neq\varnothing\implies zH=xH$. That is, the cosets form a partition of the original group. As julien noted, $x\sim y\iff y^{-1}x\in H$ is an equivalence relation on $G$. This can be seen as $x\sim y \iff \exists h\in H:hx=y$ (the cosets are the $G$ orbits of $H$ under the left translation or right translation, which is an action, which always induces a partition into orbits)


Denote $g\sim g'$ if $g^{-1}g'\in H$. Check that this defines an equivalence relation on $G$ and that equivalence classes are of the form $gH$ for some $g\in G$. Taking a set of representatives $\{g_i\;;\; i\in I\}$ for these equivalence classes yields a partition $$ G=\bigsqcup_{i\in I} g_iH. $$ Now note that $H$ is in bijection with each $g_iH$. This proves the claim for left cosets. The cardinality of $I$ is the index of $H$ in $G$, denoted by $[G:H]$. If the order of $G$ is finite, then the order of $H$ and $[G:H]$ are finite, and we get $$|G|=[G:H]|H|.$$ So the order of $H$ divides the order of $G$, which is known as Lagrange's theorem.

Do the same with $g\sim' g'$ if $g'g^{-1}\in H$ to get the right cosets. Note that the isomorphism $g\longmapsto g^{-1}$ exchanges left cosets and right cosets. In particular, $\{g_i^{-1}\;;\;i\in I\}$ is now a system of representatives for the right cosets. $$ G=\bigsqcup_{i\in I} Hg_i^{-1}. $$ It follows that we can define $[G:H]$ without ambiguity, as the cardinality of the set of left cosets, or the cardinality of the set of right cosets.


It means that $G=\cup_{g\in G} gH=\cup_{g\in G} Hg$, that the map $x\rightarrow gx$ is a bijection between $H$ and $gH$ and therefore $|gH|=|H|$. It can be deduced from this that $|G|=|G/H||H|$. A similar statement for right cosets.


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