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Definitions:

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 $H$.


Exercise:

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.

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    Sorry for this mess, and also for the bad notation. I shall come back and improve upon it, later...2012-01-29
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    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.2012-01-29
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    @BrunoStonek: Thanks very much, I will check it out.2012-01-30
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    I don't understand the qualm. It seems clear that all of the action is taking place in a fixed group $G$.2012-01-30
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    @DylanMoreland: But The projection of $H$ upon $H′/K′$ is a subgroup of $G/K'$, while that of $H'$ upon $H/K$ of $G/K$. So, if K and K' are of different size, then how can the two quotient groups in question be incident, which amounts to saying that, for each coset of the projection of $K$ upon $H'/K'$, there is one coset of the projection of $K'$ upon $H/K$ which overlap each other? Indeed, this means that there are some elements α,β, such that $αΚ=βΚ'$. If K and K' are of distinct size, how can this be true? Thanks for reading.2012-01-30
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    In that case, maybe just show me how it works, and then I shall be convinced. Thanks again.2012-01-30
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    @awllower: G/K and G/K′ are not groups. K is normal in H, not G. Subgroups L/K of H/K are sections of G, by definition. The condition is not that αΚ=βΚ′, but rather that there is exactly one β in H′ such that αΚ∩βΚ′ is non-empty. For instance G/M and M/1 are incident where G is Klein 4, and M is an order 2 subgroup, but G/M and M/1 are not incident where G is cyclic of order 4 and M is order 2.2012-01-30
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    @JackSchmidt: Might I ask why $G/M$ and $M/1$ are incident? For the coset of M in G other than M, what coset of 1 in M can have a non-empty intersection with it? Furthermore, αΚ and βΚ' are elements of the respect coset, and thus, for some coset to have elements in common, there ought to be *elements* being the same, right? Then they are the *elements*. This is exactly the bizarre statement whereby I can find nothing wrong, the essence of my problem here. Thanks for reading through this context.2012-01-30
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    In shorter terms, as the group the normal subgroup to form the quotient is one set consisting of cosets, and since the element of the coset of that group is itself a coset, αΚ in this case, for the cosets to intersect it needs that some αΚ should be the same.Is there anything wrong with this argument? Thanks for listening.2012-01-30
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    (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).2012-01-30
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    @awllower: Aha, i think I see what you are worried about. I'll fix one of their definitions.2012-01-30

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