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I would like some help in how to determine a quotient group, given two groups.

An example: The group $(\mathbb{E},+) $ in $(\mathbb{Z} , +)$.

I know the set of the quotient group is all left cosets of $( \mathbb{Z} , + )$ right? So the operation in the quotient group should be by definition: $a \mathbb{E} * b \mathbb{E} = ab \mathbb{E}$.

I'm just not sure how to piece this stuff together to easily determine the quotient group. Aren't there endless left cosets of $( \mathbb{Z} , + )$?...Or none? This is confusing me greatly..

user3200098
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    What is $\mathbb{E}$? Evens? You can just use $2 \mathbb{Z}$ since that isn't ambiguous. – Tyler Mar 05 '14 at 19:52
  • My text book just says E and I'm assuming it's even integers, yes. – user3200098 Mar 05 '14 at 19:54
  • I answered a question similar to this before (actually, i explained this exact example), let me know if it helps (see here) – Tyler Mar 05 '14 at 19:56
  • @Tyler Your answer gives me some insight into what is actually happening. So I want to take all the even integers in my example and call them the same....Which leaves me with odd integers..But that's where I stop following. You say that E+E=E which is true, but you also state that E+0=0. I don't understand that part. Also, why does it end up having only 2 elements? – user3200098 Mar 05 '14 at 20:05
  • When you group up the even integers, you separate the integers into two sets: Evens and Odds. If we think about what happens when we add evens and odds together, Even + Even = Even. Odd + Odd = Even, Even + Odd = Odd. If we think about the other groups we know, like the integers mod 2, we notice that 0 + 0 = 0, 1 + 1 = 0, and 0 + 1 = 1. Rename Even to 0, and Odd to 1, and we see it's the same group, right? It can only have two elements since there are only two ways to separate numbers by their evenness: they are or they arent. – Tyler Mar 05 '14 at 20:10
  • When I say $E + O = O$ that is the letter O (stands for ODD), not the number 0. – Tyler Mar 05 '14 at 20:11
  • I understand what you're saying but how would I write this in a ''group'' like, how would I represent this as a single group? – user3200098 Mar 05 '14 at 22:07
  • I don't know how much work you're expected to show, but: If N is a normal subgroup of G, then the index (i.e. the number of cosets) of N in G is equal to the order of the quotient group G/N. There's only one group of order two. That's all there is to it. – Tyler Mar 06 '14 at 00:31

2 Answers2

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Let's look at a slightly more "in-depth" example: Consider the subgroup $7\Bbb Z$ of $\Bbb Z$ which consists of all (positive, negative and $0$) multiples of $7$. Let's "expand" this a bit:

$7\Bbb Z = \{\dots,-21,-14,-7,0,7,14,21,28,\dots\}$

What set do we get if we add $1$ to everything in this set?

$1 + 7\Bbb Z = \{\dots,-20,-13,-6,1,8,15,22,29,\dots\}$

If we shift "over $2$" we get:

$2 + 7\Bbb Z = \{\dots,-20,-12,-5,2,9,16,23,30,\dots\}$

Now here is a curious thing: If we take something in, say $3 + 7\Bbb Z$, and add it to something in $2 + 7\Bbb Z$, we'll get something in $5 + 7\Bbb Z $. For example:

$10 + 9 = 19$ and $19$ is $5$ more than $14$.

This process "chops up" $\Bbb Z$ into $7$ pieces (each of which is "infinite") and in every piece, the numbers are all "$7$ apart".

So the "$k$" in $k + 7\Bbb Z$ measures how far to the right (right = counting up) we are from a multiple of $7$.

If you imagine the integers as an infinite string of beads, we are "wrapping the beads into a circle" so that the $7$th bead winds up in the same place as the "$0$-bead". In fact, any two beads that are a multiple of $7$ apart on our original string, wind up in the same place on our "$7$ position circle".

You've seen this before: on a clock (there it's "modulo $12$" instead of $7$, or "modulo $24$" if you use military time).

This has the effect of effectively "setting all multiples of $7$ equal to $0$". Since we know that $14$ (for example) isn't actually $0$, we don't say:

$14 = 0$, but rather, $14$ is equivalent to $0$ since it is a multiple of $7$ away from $0$ (if $6$ was "as high as we could count before starting over", $14$ would actually BE $\ 0$).

It's somewhat "magical" that all this actually WORKS, in that we don't get any contradictions or confusion this way. Part of this has to do with the fact that addition is commutative: that is, that $k + m = m+k$ for any two integers $k,m$. With a group in general, we need a more special condition on the subgroup to make this all work out (it has to be normal).

It turns out that with a general group $G$ and a subgroup $H$, that the set:

$(aH)(bH) = \{(ah)(bh'): h,h' \in H\}$ doesn't always equal some other coset $kH$, so subgroups for which this DOES happen are "special".

If $G$ is abelian, of course, then $(ah)(bh') = a(hb)h' = a(bh)h' = (ab)(hh')$ and this DOES happen. So quotient groups of abelian groups are "easier to understand".

In this special case of $G = \Bbb Z$ and $H = ${even integers}, this process is called "the arithmetic of parity" (reasoning by evens and odds). This "trick" (reducing mod $2$) turns an "infinite number of cases" to just TWO cases, and often that is all we need.

David Wheeler
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If you mean the quotient of the group $(\mathbb{Z},+)$ inside $(\mathbb{R},+)$: indeed there are infinitely many cosets: for each $x \in [0,1)$, the set $x + \mathbb{Z} = \{x+z : z \in \mathbb{Z}\}$ is a coset. These cosets are all different, and these are all the cosets.

The resulting quotient group is the group $([0,1),+)$ where addition works modulo $1$ (we identify each cosets with its generator in $[0,1)$). This is the circle group $\mathbb{R}/\mathbb{Z}$.

If $\mathbb{E}$ is the set of even integers: there are two cosets: the set of odd integers and the set of even integers. The resulting quotient group is the group $\mathbb{Z}/2\mathbb{Z}$ with two elements.

Note that you're working in additive groups; the operation on cosets is $(a+\mathbb{Z}) + (b + \mathbb{Z}) = a+b+\mathbb{Z}$. When $a$ is odd, $a+\mathbb{Z}$ is the set of odd integers; when $a$ is even, $a + \mathbb{Z}$ is the set of even integers.

Denote the cosets by $X$ (even integers) and $Y$ (odd integers). Then the group operation yields $X+X=X$, $X+Y=Y$ and $Y+Y=X$.

user133281
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