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Why do we take normal subgroups when we define a quotient group?
Charles S.
former mathematician, current patent lawyer · Upvoted by
Alon Amit
, Ph.D. in Mathematics. and
Erik Bergland
, PhD Mathematics, Brown University (2024)Author has 7.2K answers and 51.8M
answer views · 5y ·

Because otherwise it doesn’t work.

Let’s pretend we didn’t know that. Let GG be a group and HH a
not-necessarily-normal subgroup. We want to define G/HG/H. To me, the most
intuitive characterization of G/HG/H is that you don’t care about “multiples of
stuff in HH.” This is borrowed straight from the concept of modular arithmetic,
so it’s going to be my guiding principle in trying to define G/HG/H.

The other guiding principle is that, since we’re in group theory land, I also
want G/HG/H to be a group. (After all, the integers mod 5 isn’t very useful if
you couldn’t do arithmetic mod 5.)

Okay, let’s proceed normally: define G/HG/H to be the left cosets of HH; that
is, {gH}{gH} for g∈Gg∈G. We want the natural group operation: if g1Hg1H and
g2Hg2H are cosets, the natural operation is defined by
(g1H)(g2H)=g1g2H(g1H)(g2H)=g1g2H.

Looks right, but it actually doesn’t make sense. Why? Because there’s more than
one way to represent g1Hg1H, at least if HH is not just the zero subgroup. In
more detail, it might be that f1H=g1Hf1H=g1H, with f1≠g1f1≠g1. By way of example
in modular arithmetic, the integers 1, 4, and 10 “represent” the same number mod
3. In the case of modular arithmetic, we have the benefit of the integers being
ordered, so we can all agree to pick a standard representative: the one between
0 and nn in the integers modulo nn. But with a generic subgroup HH of a generic
subgroup GG, there’s nothing special going on to help us pick a standard
representative.

Fine, so we have two representatives: f1H=g1Hf1H=g1H, even though f1≠g1f1≠g1.
Let’s also say f2H=g2Hf2H=g2H, with f2≠g2f2≠g2. If our multiplication in G/HG/H
is going to be well-defined, we need f1f2H=g1g2Hf1f2H=g1g2H. This last statement
is actually very important, so don’t gloss over it. If it’s clear, skip the next
paragraph. Otherwise…

--------------------------------------------------------------------------------

By way of yet another analogy, the same work of fiction can be printed in a book
in different ways, and in particular having different page numbers. So if you
try to define a “page count” function on works of fiction, it’s not well
defined. Hamlet might be printed with 200 pages by one publisher, and 3 pages
(with very small fonts, or very large pages) by another. The “page count”
function is perfectly defined over the set of books, but not on abstract works
of fiction. You can write the same work of fiction down different ways.

--------------------------------------------------------------------------------

Okay, let’s try to prove our operation on cosets is well-defined. Since
f1H=g1Hf1H=g1H, there’s an h1∈Hh1∈H with f1=g1h1f1=g1h1. (Exercise: I could
write f1h11=g1h12f1h11=g1h12, or something else equally notationally horrific.
Why don’t I need to?). Similarly, there’s an h2∈Hh2∈H with f2=g2h2f2=g2h2. Now
the big moment:

f1f2=g1h1g2h2f1f2=g1h1g2h2.

We want to be able to write f1f2=g1g2hf1f2=g1g2h for some h∈Hh∈H. That would
mean our operation in G/HG/H is well-defined. But without some extra tools, we
can’t ever “kill” the h1h1 floating around between the g1g1 and g2g2 terms
above.

But what if H were normal?

Then gH=HggH=Hg for any g∈Gg∈G. In other words, you could find an h3h3 so that
h1g2=g2h3h1g2=g2h3. Plug that into the “big moment” equation, and you get

f1f2=g1(h1g2)h2=g1(g2h3)h2f1f2=g1(h1g2)h2=g1(g2h3)h2.

Now you can collapse all the hihi-terms on the right to a single hh, giving us
what we want:

f1f2=g1g2hf1f2=g1g2h

for some h∈Hh∈H.

Success!


3.8K views ·
View upvotes
·
1 of 4 answers
Comments
Vance Faber
· 5y


By choosing g_1 and g_2 to be inverses of each other you can prove that the
cosets form a group if and only if H is normal.

View 3 other answers to this question
About the Author
Charles S.
Former mathematician, current patent lawyer
Studied at University of California, Santa Barbara
Lives in Cambridge, MA
51.8M content views551.4K this month
Top Writer2018
Active in 1 Space
Joined October 2013
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