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How do "Identify" Quotient Groups with Another Standard Group (Algebra Self Study)

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Edited by Zafar Bakhromov, Wednesday 1 April 2026 at 13:44

A Quotient Group cap g solidus cap n is the set of cosets of cap n in cap g , with a law of composition that makes  it into a group. The whole idea is that, if we write the law of composition on cap g as " plus ", then we want to view two elements theta and theta plus cap n that "differ by a multiple of cap n " as the same element in cap g solidus cap n .

While self-studying the book "Algebra 2.e." by Michael Artin, I found it somewhat difficult to grasp how to analyze the structure of a quotient group. Here I will summarize what I've learned through several exercises.

Exercise:

Let cap g be the group of upper triangular real matrices matrix row 1column 1 ab row 2column 1 zero d , with a and d different from zero. For each of the following subsets, determine whether or not cap s is a subgroup and whether or not cap s is a normal subgroup. If cap s is a normal subgroup, identify the quotient group cap g solidus cap s .

(i) cap s is the subset defined by b equals zero .

(ii) cap s is the subset defined by d equals one .

(iii) cap s is the subset defined by a equals d .

Source: "Algebra" 2.e., Michael Artin, p.75

My Solution:

(i) cap s is the set of two multiplication two diagonal matrices, with nonzero entries a and d on the diagonal. This forms a normal subgroup of cap g , but we omit the proof to keep our focus on quotient groups.

Since cap g solidus cap s colon equals left curly bracket g times cap s vertical line g element of cap g right curly bracket , an element  g times s of g times cap s looks like:

equation sequence part 1 g times s equals part 2 matrix row 1column 1 ab row 2column 1 zero d times matrix row 1column 1 x zero row 2column 1 zero y equals part 3 matrix row 1column 1 times times ax times times by row 2column 1 zero times times dy

and we need to choose a representative element for each g times cap s to analyze cap g solidus cap s . With the entries x and y of s free to choose for any g , we may pick x colon equals one divided by a and y colon equals one divided by d to make multiplication easier. Hence, denoting a representative element of g times cap s as g macron , we get:

equation sequence part 1 g times cap s equals part 2 g macron equals part 3 matrix row 1column 1 one bd row 2column 1 01 equals part 4 matrix row 1column 1 one b prime row 2column 1 01

where we have defined b super prime equals b divided by d . Thus, we have reduced g times cap s element of cap g solidus cap s into workable form.

Now, to analyze cap g solidus cap s , we have to multiply two of its elements. Taking g macron comma h macron element of cap g solidus cap s , we have that:

equation sequence part 1 g macron times h macron equals part 2 matrix row 1column 1 one b prime row 2column 1 01 times matrix row 1column 1 one e prime row 2column 1 01 equals part 3 matrix row 1column 1 one plus plus b prime e prime row 2column 1 01

Hence, all multiplication does in cap g solidus cap s is that it adds the entries in the top right corners of g macron and h macron , so it really looks like the addition of real numbers.

We are now ready to identify the quotient group cap g solidus cap s with a standard group. Let f colon cap g solidus cap s right arrow double-struck cap r super plus comma matrix row 1column 1 one b prime row 2column 1 01 right arrow from bar b super prime , where double-struck cap r super plus denotes the additive group of real numbers. Then, f is an isomorphism. The bijectivity of f is given by the fact that  b super prime can be any real number (including 0), and f is a homomorphism because:

equation sequence part 1 f of matrix row 1column 1 one b prime row 2column 1 01 times matrix row 1column 1 one e prime row 2column 1 01 equals part 2 f of matrix row 1column 1 one plus plus b prime e prime row 2column 1 01 equals part 3 b super prime plus e super prime equals part 4 f of matrix row 1column 1 one b prime row 2column 1 01 plus f of matrix row 1column 1 one e prime row 2column 1 01

Hence, cap g solidus cap s is isomorphic to double-struck cap r super plus .

(ii) We similarly have that cap s is a normal subgroup of cap g . For some particular g element of cap g , an element g times s of g times cap s looks like:

equation sequence part 1 g times s equals part 2 matrix row 1column 1 ab row 2column 1 zero d times matrix row 1column 1 xy row 2column 1 01 equals part 3 matrix row 1column 1 times times ax plus plus times times ayb row 2column 1 zero d

With the entries x comma y of s free to choose, we may take x colon equals one solidus a and y colon equals negative b solidus a , in which case we get that:

g macron equals matrix row 1column 1 10 row 2column 1 zero d

so

equation sequence part 1 g macron times h macron equals part 2 matrix row 1column 1 10 row 2column 1 zero d times matrix row 1column 1 10 row 2column 1 zero f equals part 3 matrix row 1column 1 10 row 2column 1 zero times times df

Thus, by similar logic as before, cap g solidus cap s approximately equals double-struck cap r super multiplication , where double-struck cap r super multiplication is the multiplicative group of nonzero real numbers.

(iii) I'm getting a little tired of typesetting this so we're just going to note that cap g solidus cap s approximately equals double-struck cap r super multiplication by similar reasoning as in (ii).

Conclusion:

It took me about a week of working through examples like these to really grasp what a quotient group is doing. The key insight, for me, was not just the definition, but learning how to choose useful representatives of cosets. By simplifying each coset to a canonical form, the group operation in cap g solidus cap s  becomes explicit and often reveals a familiar structure. In this way, quotient groups stop feeling like abstract sets of cosets and instead become concrete objects that capture exactly what remains of a group after “modding out” a chosen symmetry.

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