Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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


If $K$ is a normal subgroup of a subgroup $H$ of $G$, then $H/K$ is referred to as a section of $G$.

Two sections $H/K$, $H'/K'$ of the group $G$ are said incidents if any coset in $H/K$ has an intersection with one unique coset in $H'/K'$ that is not empty, and the converse.

Let $L/M$ be one section of $G$ and $H$ a subgroup of $G$. We define the projection of $H$ upon $L/M$ to be the subset of $L/M$ that consists of cosets of $M$ in $L$ overlapping with $H$.


i. Incident sections are isomorphic.

ii. Let $N$ be one normal subgroup of $G$, $H$ one subgroup of $G$. Then $HN/N$ and $H/(H \cap N)$ are incident.

iii. Conditions being as the definition iii., the projection of $H$ upon $L/M$ is the subgroup $(L \cap H)M/M$ of $L/M$.

iv. The projection of $K'$ upon $H/K$ is a normal subgroup of the projection of $H'$ upon $H/K$. Then this quotient group is considered the projection of $H'/K'$ upon $H/K$.

v. The projections, respectively, of $H/K$ upon $H'/K'$ and of $H'/K'$ upon $H/K$ are incident.

The Final Result:

Let $H$, $H'$ be subgroups of $G$, $K$ one normal subgroup of $H$, $K'$ that of $H'$. Then $ (H \cap {H'})K/(H\cap{K'})K $ is isomorphic to $(H\cap{H'})K'/(K\cap{H'})K'$.

It is taken from the book: Groups and representations, by JL Alperin, and Rowen B Bell. Here $A\cap B$ is the intersection of them.
This was first proved by Zassebhaus, at the aureate age of 21, whereupon leaving the name of the lemma of Zassenhaus, the fourth isomorphism theorem, or the butterfly lemma, owing to the shape of its inclusion diagram of involved subgroups.

Question: I can without confronting many difficulties solve the first four exercises, while the last one still puzzles me. In effect, what confuses me is that, in order to consider two sections as incident, shall we not first be sure that they are sections with respect to the same group $G$? I cannot at this juncture even assume that the fifth exercise is true...
Thanks for any hint.

share|cite|improve this question
Sorry for this mess, and also for the bad notation. I shall come back and improve upon it, later... – awllower Jan 29 '12 at 11:19
You may want to compare the proof outlined by the exercises with one full proof in another algebra book. I remember studying the proof from Rotman's Advanced Modern Algebra (indeed: it's on page 279). The lemma is useful in proving the Jordan-Hölder theorem. – lentic catachresis Jan 29 '12 at 13:26
@BrunoStonek: Thanks very much, I will check it out. – awllower Jan 30 '12 at 3:04
I don't understand the qualm. It seems clear that all of the action is taking place in a fixed group $G$. – Dylan Moreland Jan 30 '12 at 3:27
(small mistake corrected) If G={1,x,y,xy} is not cyclic, M={1,x}, and N={1,y}, then G/M = { {1,x}, {y,xy} } is incident to N/1 = { {1}, {y} } because {1,x} and {1} intersect at 1, and {y} and {y,xy} intersect at y, but other cosets (like {1,x} and {y}) do not intersect. In my previous comment, I miswrote, since G/M and M/1 are never incident in any non-identity group (if M={1} and g≠1, then gM intersects every coset of {1} in M in the empty set; if m in M and m ≠ 1, then 1M intersects both m1={m} and 11={1}, so also not incident). – Jack Schmidt Jan 30 '12 at 16:06

Here is an example, since the comments are getting crowded:

Let $G = \{ 1, x, y, xy \}$ be a Klein 4-group. The sections of G are precisely:

  • $X/1 = \{ \{ 1 \}, \{ x \} \}$
  • $Y/1 = \{ \{ 1 \}, \{ y \} \}$
  • $Z/1 = \{ \{ 1 \}, \{ xy \} \}$
  • $G/X = \{ \{ 1, x \}, \{ y, xy \} \}$
  • $G/Y = \{ \{ 1, y \}, \{ x, xy \} \}$
  • $G/Z = \{ \{ 1, xy \}, \{ x, y \} \}$
  • $G/1 = \{ \{ 1 \}, \{ x \}, \{ y \}, \{ x, y \} \}$
  • $1/1 = \{ \{ 1 \} \}$
  • $X/X = \{ \{ 1, x \} \}$
  • $Y/Y = \{ \{ 1, y \} \}$
  • $Z/Z = \{ \{ 1, xy \} \}$
  • $G/G = \{ \{ 1, x, y, xy \} \}$

The following (unordered) pairs of sections are incident:

  • $(X/1,G/Y)$ since $\{1\} \cap \{1,y\}= \{1\}$, $\{1\} \cap \{x,xy\} = \varnothing$, $\{x\} \cap \{1,y\} = \varnothing$ and $\{x\} \cap \{x,xy\} = \{x\}$ (so that each coset in $X/1$ intersects exactly one coset in $G/Y$)
  • $(X/1,G/Z)$
  • $(Y/1,G/X)$
  • $(Y/1,G/Z)$
  • $(Z/1,G/X)$
  • $(Z/1,G/Y)$
  • all the $H/H$ are incident, since there is only one coset per section, and all the cosets contain the identity 1 of G

This is all (I believe). For instance $G/X$ and $G/Y$ are not incident, since $\{1,x\}$ intersects both $\{1,y\}$ and $\{x,xy\}$. Of course $G/X \to Z/1 \to G/Y$ shows the two sections are "connected" and so isomorphic, but they are not incident as in directly next to each other.

Similarly, $G/X \to Y/1 \to G/Z \to X/1$ shows that $G/X$ and $X/1$ are connected and isomorphic, but obviously not incident (sorry!) since $\{1,x\}$ intersects both $\{1\}$ and $\{x\}$ non-emptily.

share|cite|improve this answer

Definition: If $H/K$ is a section of $G$ and $H'/K'$ is a section of $G$, then the projection of $H/K$ onto $H'/K'$ is the set of all cosets $hK'$ where h is in both H and H

Lemma: The projection of a section of G onto a section of G is a section of G.

Proof: The projection of $H/K$ onto $H'/K'$ is precisely the section $((H\cap H')K')/K'$. $\square$

The original question only defines the projection of a subgroup, and obviously that subgroup needs to be a subgroup of G not $H/K$ for anything to make sense, so @awllower was understandably concerned. Of course the projection of $H/K$ is the same as the projection of $H$, so perhaps the author felt no need to define it.

share|cite|improve this answer
Actually part (iv) makes the same definition, so the exercise seems fine as stated. – Jack Schmidt Jan 30 '12 at 18:13
Sorry, but the definition for the projection of a section upon another is not like what you stated. In effect, it is the quotient group of the projection of $H'$ on $H/K$ by the projection of $K'$ on $H/K$. Thus it will consist of cosets of sets of cosets. This is what I cannot understand. Per chance I am making a wrong announcement here? But this is what I perceived from the book I mentioned, and not until I return to the school can I get my hands on another book(big Rotman). Therefore I am still looking forward to any clarification, thanks. – awllower Feb 3 '12 at 9:21

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.