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Let $G$ be a finite group and $H$ a subgroup. Is it true that a set of right coset representatives of $H$ is also set of left coset representatives of $H$?

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No. There is a fun salvage though: there exists a set of right coset representatives which is also a set of left coset representatives. This is a corollary of Hall's marriage theorem (!). – Qiaochu Yuan Aug 2 '12 at 23:03
It is a fun exercise to find such a system for $H = S_m < S_n = G$ with $m < n$. A more combinatorial question is to count/estimate the number of all such systems. – Danny Nguyen May 30 '14 at 6:54
up vote 6 down vote accepted

Not every left transversal is also a right transversal. The Group Properties Wiki has a list of subgroup properties that are stronger than "having [at least one] left transversal that is also a right transversal"; among these is normality as William notes.

However the left coset representatives' multiplicative inverses form a right transversal, because

$$\begin{array}{c l}xH=yH & \iff y^{-1}xH=H \\ & \iff y^{-1}x\in H \\ & \iff (y^{-1}x)^{-1}=x^{-1}y\in H \\ & \iff Hx^{-1}y=H \\ & \iff Hx^{-1}=Hy^{-1}. \end{array}$$

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GroupProps also says that "every left transversal is a right transversal if and only if it is a normal subgroup," but I don't immediately see why "only if." This is the substance of my comment on William's answer too. – anon Aug 2 '12 at 22:41
Although speaking without having a solid criteria is a worst job in Math, I saw somewhere if a left transversal be a right transversal implies that $H$ is normal in $G$. I lost the reference but I am sure I saw this claim. Sorry. – Babak S. Aug 2 '12 at 23:11
@BabakSorouh That can't be right. As Steve notes in a comment to the deleted answer, every subgroup of a finite group has some transversal that is also a right transversal. (And of course, not every subgroup of finite groups is normal!) One needs to differentiate between at least one and all in order to correctly entail normality for left/right transversality. – anon Aug 2 '12 at 23:17
Thanks. I see that. :) – Babak S. Aug 2 '12 at 23:22

This answer is mostly to give a nice bibliography for this problem of finding common transversals. I was going to include a proof, but all the proofs were combinatorial and my attempts at making them sound interesting failed.

Exists: yes, All: no

If $H \leq G$ and $K \leq G$ are two subgroups of the same finite index $[G:H]=[G:K]$, then they have a “common transversal”: a set $t_i \in G$ such that $H t_i \neq H t_j$ and $t_i K \neq t_j K$ if $i \neq j$ and such that $G = \cup H t_i = \cup t_i K$. Taking $H=K$ gives an answer to the original question, interpreting "a" as "there exists" rather than "for all". I discuss the publication history of this theorem in the next section.

One can easily check that $\{ (), (2,3), (1,2,3) \}$ is only a one-sided set of coset representatives of $\{(),(1,2)\}$ in the symmetric group $\{ (), (1,2), (2,3), (1,3), (1,2,3), (1,3,2) \}$.


The earliest proof is due to Miller (1910) with another similar proof in Chapman (1913). The same proof was used for a stronger result in Scorza (1927). Van der Waerden (1927) proved a combinatorial precursor to the Marriage theorem of Hall (1935) and explicitly cites Miller (1910) as his motivation. A proof is presented in textbook form in Zassenhaus (1937), theorem 3, page 12; I believe this is van der Waerden's proof, but Zassenhaus credits it to Willi Maak. Shü (1941) generalizes Scorza's result to finite index subgroups and gives an example to show the result is not true for a general subgroup. Ore (1958) continues this investigation. Hall (1967) uses the marriage theorem to prove the theorem in the finite index case as theorem 5.1.7 on page 55. Alonso (1972) proves less, but gives more examples and uses more primitive versions, somewhat similar to Hall (1967). Applegate–Onishi (1977) proves the result for continuous sections of profinite groups.


  • Miller, G. A. “On a method due to Galois.” Quarterly Journal of Mathematics 41, (1910), 382-384. JFM41.174.1
  • Chapman, H. W. “A note on the elementary theory of groups of finite order.” Messenger of Mathematics (2) 42 (1913), 132-134; correction 43 (1914), 85. JFM44.168.3 (Correction in next volume)
  • Scorza, G. “A proposito di un teorema del Chapman.” Bollettino della Unione Matemàtica Italiana 6, 1-6 (1927). JFM53.105.5
  • van der Waerden, B. L. “Ein Satz über Klasseneinteilungen von endlichen Mengen.” Abhandlungen Hamburg 5, 185-187 (1927). JFM53.171.2 DOI10.1007/BF02952519
  • Hall, P. “On representatives of subsets.” Journ. London Math. Soc. 10, 26-30 (1935). JFM61.67.1 ZBL10.345.3 DOI:10.1112/jlms/s1-10.37.26
  • Zassenhaus, Hans. Lehrbuch der Gruppentheorie. (Hamburg. Math. Einzelschriften 21) Leipzig, Berlin: B. G. Teubner. VI, 152 S. (1937). JFM63.58.3 ZBL18.9.1 MR30947 MR91275
  • Shü, Shien-siu. “On the common representative system of residue classes of infinite groups.” J. London Math. Soc. 16, (1941). 101–104. MR4624 DOI:10.1112/jlms/s1-16.2.101
  • Ore, Oystein. “On coset representatives in groups.” Proc. Amer. Math. Soc. 9 (1958) 665–670. MR100639 DOI:10.2307/2033229
  • Hall, Marshall, Jr. Combinatorial theory. Blaisdell Publishing Co. Ginn and Co., Waltham, Mass.-Toronto, Ont.-London (1967) x+310 pp. MR224481 MR840216
  • Alonso, James. “Representatives for cosets.” Amer. Math. Monthly 79 (1972), 886–890. MR315004 DOI:10.2307/2317669
  • Appelgate, H.; Onishi, H. “Coincident pairs of continuous sections in profinite groups.” Proc. Amer. Math. Soc. 63 (1977), no. 2, 217–220. MR442099 DOI:10.2307/2041792
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No this not true unless $H$ is a normal subgroup. Suppose that $H$ is normal. Then for all $a \in G$, $aHa^{-1} = H$ so $aH = Ha$. Suppose that the set of left and right cosets are the same. Since cosets are partitions of $G$, $aH = Ha$. Hence $aHa^{-1} = H$. Since this holds for all $a \in G$, $H$ is a normal subgroup.

However for any subgroup $H$ of $G$ it is true that the set of left cosets and right cosets are in bijection. Namely $\Phi(aH) = Ha^{-1}$.

Let $R$ and $L$ be the set of right or right and left cosets. Above I showed that $R = L$ if and only $H$ is normal .

Now suppose that $H$ is not normal. So $R \neq L$. Hence there is a right coset $\sigma$ that is not a left coset. Since you assumed that the groups are finite, the size of each left and right cosets are equal. Hence if $\sigma$ is not any left coset, then $\sigma$ must intersect two left cosets $\tau_1$ and $\tau_2$. Let $a_1 \in \sigma \cap \tau_1$ and $a_2 \in \sigma \cap \tau_2$. Now form a set of representatives $K$ for left cosets, where $a_1$ and $a_2$ are choosen representative for $\tau_1$ and $\tau_2$ respectively. Now $K$ is not a set of representative for right cosets since $a_1$ and $a_2$ are representative for the same right coset. So some right coset must have no representative.

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@anon I guess it is more precise to say "suppose the set of left and right cosets are the same". Since cosets are equivalence classes, the left coset containing $a$ is $aH$ and the right coset containing $a$ is $Ha$. If the set of left and right cosets are the same and since cosets partition $G$, you must have $aH = Ha$. Nowhere does this require $H$ to be normal. – William Aug 2 '12 at 22:39
@anon I think I understand your concern. "Suppose the left and right cosets are the same" means the set of left coset and the set of right cosets are the same. There the set of left cosets is a just a set of subsets of $G$. Simiarly the set of right cosets are just a set of subsets of $G$. Are these two elements of the power set of the powerset of $G$, the same? It is just a question about set equality. – William Aug 2 '12 at 22:46
@anon I don't understand what you are saying. Let $R$ and $L$ be the set of right and left cosets, respectively. Since cosets form a partition, there exist a left coset $\sigma \in L$ such that $a \in \sigma$. Since $R = L$, there exists a right coset $\tau \in R$ such that $\sigma = \tau$. Since $\sigma$ is a left coset containing $a$, you know that $\sigma = aH$. Similarly $\tau = Ha$. So $aH = \sigma = \tau = Ha$. So $aHa^{-1} = H$. $a$ was arbitrary so $H$ is normal. Is this any better? – William Aug 2 '12 at 22:54
This is not correct: In any finite group, one can find transversals that work for both left and right cosets, yet not every subgroup is normal. – user641 Aug 2 '12 at 23:04
@William: No, the OP asked if every left transversal was a right transversal, not if every left coset partition was a right coset partition. You speak of the latter. – anon Aug 2 '12 at 23:06

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