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Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In 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 More answers from Charles S. 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