Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove the following: If $H$ is a subgroup of finite index in a group $G$, and $K$ is a subgroup of $G$ containing $H$, then $K$ is of finite index in $G$ and $[G:H] = [G:K][K:H]$.

So this is basically a bijective proof? The number of cosets of $H$ in $G$ equals the number of cosets of $K$ in $G$ times the number of cosets of $H$ in $K$ by the multiplication principle?

Reference: Fraleigh p. 103 Question 10.35 in A First Course in Abstract Algebra

share|cite|improve this question
I'm not sure exactly what "the multiplication principle" is, but yes, the proof goes as you describe. They key is that each of the $[G:K]$ cosets of $K$ breaks into $[K:H]$ cosets of $H$ (this is worth writing down carefully) and now we, um, multiply. Note that if you're willing to do cardinal arithmetic, no finiteness assumptions are needed here. – Pete L. Clark Oct 17 '10 at 4:44
up vote 5 down vote accepted

In fact, the result holds even if you do not assume that we are dealing with finite index, provided you take the equality to be an equality of cardinalities. Just follow the idea Pete Clark suggests, but without assuming that the number of cosets is finite in either case. And remember that the number of distinct cosets of a subgroup $B$ in a group $A$ is equal, by definition, to the index of $B$ in $A$, $[A:B]$.

As to whether the $x_i$ are "coset representatives for $H$", remember that for a collection of elements to be "a set of coset representatives" we require that if $x_iH = x_jH$, then $i=j$, and that for every $g\in G$ there exists $i$ such that $gH = x_iH$ (thinking left cosets; right cosets work the same way). So $x_1,\ldots,x_n$ will not generally be "a coset representative" nor a set of coset representatives, but that is immaterial. Simply show that if $\{x_i\}_{i\in I}$ is a set of coset representatives for $K$ in $G$ (meaning, (i) for every $g\in G$ there exists $i\in I$ such that $gK = x_iK$; and (ii) for all $i,i'\in I$, if $x_iK = x_{i'}K$ then $i=i'$), and $\{y_j\}_{j\in J}$ is a set of coset representatives for $H$ in $K$ (meaning, (1) for every $k\in K$ there exists $j\in J$ such that $kH = y_jH$; and (2) for all $j,j'\in J$, if $y_jH = y_{j'}H$, then $j=j'$); then it follows that $\{x_iy_j\}_{(i,j)\in I\times J}$ is a set of coset representatives for $H$ in $G$ (meaning, you need to prove that (I) for all $g\in G$ there exists $(i,j)\in I\times J$ such that $gH = (x_iy_j)H$; and (II) for all $(i,j),(i',j')\in I\times J$, if $(x_iy_j)H = (x_{i'}y_{j'})H$, then $(i,j)=(i',j')$).

A fact that will no doubt be useful is to remember that for any group $A$ and any subgroup $B$ of $A$, $cB = dB$ if and only if $cB\cap dB\neq\emptyset$.

share|cite|improve this answer
The infinite case comes up in Dummit & Foote... Where did the $x_i$ come from? And where did "coset representative" come from? – Altar Ego Sep 9 '11 at 12:27
@Altar Ego: As I mention in the first paragraph, I was following in the idea that Pete Clark had suggested in a now-deleted answer,which began by selecting $x_i$ to be coset representatives for $K$ in $G$, $y_i$ to be coset representatives for $H$ in $K$, and suggested showing $\{x_iy_j\}$ were coset representatives for $H$ in $G$. The OP asked a question in comments there, and my paragraph was answering that in the general case. – Arturo Magidin Sep 9 '11 at 13:12
Thanks for the clarification. I usually feel lost, much more so when there's a deleted answer! – Altar Ego Sep 9 '11 at 16:31
@Altar Ego: Once you get enough reputation, you'll be able to "see" deleted answers. – Arturo Magidin Sep 9 '11 at 17:00
Can't wait! I've already got a gold badge :) – Altar Ego Sep 9 '11 at 17:32

The canonical map $G/H \to G/K$ is surjective. The fiber of $gK$ is $\{gkH : k \in K\}$, which can be identified with $K/H$.

share|cite|improve this answer

While studying today, I think I did a very similar exercise. It is presented as:

Being $G$ a finite group and $H \leq K \leq G$, prove that $|G:H| = |G:K|\cdot|K:H|$.

I answered the following:

$|G:K|\cdot|K:H| = \frac{|G|}{|K|}\cdot\frac{|K|}{|H|} = \frac{|G|\cdot|K|}{|K|\cdot|H|} = |G:H|$

Is it the same thing or am I mistaking anything?

Thank you for the attention.

(I know this isn't an answer to your question, and if you find this post spam, please let me know.)

share|cite|improve this answer
For finite groups, assuming you already know that $|G:H|=\frac{|G|}{|H|}$ (whether because that's how you defined $|G:H|$ or because you proved it), then this works; and if what you write as $|G:H|$ is $[G:H]$, the index of $H$ in $G$ (the number of left cosets of $H$ in $G$), then yes, it is the same thing (under the finiteness assumption). – Arturo Magidin Feb 17 '11 at 22:35
@Arturo Magidin: Yes, with $|X|$ i mean $[X]$. I weren't aware this two notations meant the same, so I used the one I am used to. Thanks for your confirmation! Very much appreciated! – Marla Feb 17 '11 at 22:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.