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I'm having a hard time thinking about factor groups. I just don't understand what notation like $\mathbb{Z}_{60}/\langle 12 \rangle$ means. Furthermore, when asked about giving the order $26 + \langle 12 \rangle$ in $\mathbb{Z}_{60}/\langle 12 \rangle$, I don't know how to picture it.

I've seen the solution that $26 + \langle 12 \rangle = 2 + \langle 12 \rangle$. The cyclic subgroup of $2 + \langle 12 \rangle$ is $\{\langle 12 \rangle, 2 + \langle 12 \rangle, 4 + \langle 12 \rangle, 6 + \langle 12 \rangle, 8 + \langle 12 \rangle, 10 + \langle 12 \rangle\},$ hence we have an order of 6, but I don't see the relation between that and $\mathbb{Z}_{60}/\langle 12 \rangle$. I can't seem to visualize it.

If I could see another solution with $\mathbb{Z}_{60}/\langle 12 \rangle$ formulaically broken down into its components, I think I would have a much easier time with all of this.

I understand $\mathbb{Z}_{60}$ is quite large, so you could use a much simpler case. Thank you very much.

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Your second equality on that line is wrong. On one side you have a subgroup, and on the other you have just one element (ie, one coset). –  Tobias Kildetoft Oct 8 '13 at 17:54
    
You mean the $2 + \langle 12 \rangle$? –  David Oct 8 '13 at 17:55
    
Yes, that is one coset, ie one element in the quotient group. –  Tobias Kildetoft Oct 8 '13 at 17:57
    
I would suggest that you read Abstract Algebra by Dummit and Foote. The chapter on Homomorphisms explains it really well, with pictures. –  Vishal Oct 8 '13 at 17:59
    
@TobiasKildetoft, I thought $26 + \langle 12 \rangle$ is equivalent to $2 + \langle 12 \rangle$ because $26 \equiv 2 \mod 12$. –  David Oct 8 '13 at 18:03

3 Answers 3

Here's some intuition on the meaning of a factor group.

Reference: http://www.math.toronto.edu/lgoldmak/301F09/LS10-27.pdf

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Reference: http://cdsmith.wordpress.com/2009/04/11/needing-intuition-in-maths-one-example/

Most people learning abstract algebra, as far as I can tell, have no idea why homomorphisms and factor groups are sensible things to think about. They quickly come to understand the idea of a group, and enough varied examples are usually given that they can see how the idea of a group applies to a number of things. They quickly come to terms with subgroups, though the idea looks rather trivial to them. Then you get to homomorphisms and factor groups; at this point, most classes run out of intuition and just jump in for some unmotivated mathematical constructions.

I’m not quite sure why this is, honestly. Anyone with the slightest modicum of mathematical curiosity probably has thoughtn about factor groups since they were seven or eight years old. In the context of integers and addition, most children realize on their own (whether it’s taught to them or not) that the sum of two even numbers, or of two odd numbers, is even, while the sum of an even number and an odd number is odd. This is, of course, a factor group. Students who are presented with the mathematical definition of a factor group should first have, in their set of mental tools, this simple intuitive definition:

Factor Group: For any group (G,*), a factor group is a group that is obtained by being sufficiently sleep-deprived (or perhaps drunk, depending on the university) that one can’t tell the difference between some members of the original group, and then trying to write down a group table.

Of course, one then goes on to point out that sometimes this works, but sometimes it doesn’t. If one looks at the integers and only sees “even” or “odd”, then it works. If one looks at the integers and only sees “negative” or “non-negative”, then it doesn’t work, since the sum of a negative number and a positive number could be either negative or positive. It then becomes natural to ask when it works, and when it doesn’t. This provides a justification, then, for nailing down the abstract definitions, defining normal subgroups, proving that the factor group is well-defined when modding out a normal subgroup, and so on. First, though, the student needs to be convinced that these are natural things to think about.

Speaking of defining normal subgroups, it is really inexcusable how many students have never even noticed the close relationship between normal subgroups and commutativity. Sure, everyone knows that all subgroups of an abelian group are normal; but this seems to be treated as a sort of occasionally useful curiosity. Few students are even exposed to the simple fact that normality of subgroups is inextricably entwined in the degree to which the subgroup commutes with the surrounding group.

Reference: http://scienceblogs.com/goodmath/2007/12/27/the-meaning-of-division-quotie/

A quotient group is a shining example of the beauty of abstract algebra. We’ve abstracted away from numbers to these crazy group things, and one reward is that we can see what division really means. It’s more than just a simple bit of arithmetic: division is a way of describing a fundamental structural relationship that pervades mathematics.

So what is division all about?

Suppose you want to divide 50 by 5. What you’re really doing is saying you’ve got a collection of 50 indistinguishable things, and you want to break it into a 5 indistinguishable collections. Since you started with 50 indistinguishable things, that means that you’ll end up with 5 sets of 10 things.

That’s pretty simple, right? Now, suppose that we’re not talking about simple numbers. Instead we want to work in terms of groups. Can we take that basic concept of division, and find some meaningful way of applying it to groups?

Well, first, we need to somehow talk about division in a way that doesn’t involve numbers. We start with a group – that is, a collection of objects with some kind of meaningful structure. What can we divide it by? A group with a similar structure – in fact, a group with the same structure: a subgroup But not just any subgroup: it’s got to be a normal subgroup, because that’s the kind of subgroup that properly preserves the structure of the group.

So what happens when we divide a group by one of its normal subgroups? We partition the group into a new group, where the elements of the new group are formed from subsets of the elements of the original group. It’s the same idea as simple integer division described up above, except that we want to preserve the group structure, so the result is going to be a group.

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You should always think of quotients as being equivalence relations. If $\sim$ is an equivalence relation on $A$, then $A/\sim$ is the set of equivalence classes. You can visualize this by imagining that you have just taken $A$, but somehow identified (or, to use a visual word, glued) $a$ and $b$ whenever $a\sim b$.

What you are doing here is strengthening equality, or forcing some things to equal some other things. For example, maybe I'm working with the set $\{x,y\}$, where $x$ represents my brother and $y$ represents my roommate. But then, maybe I move in with my brother—now this is the wrong set to use, and I should use $\{x,y\}/\sim$, where $x\sim y$, because I now want to think of my brother and my roommate as two different people.

A more complicated example would be the group $\mathbb{Z}/24\mathbb{Z}$. The equivalence relation here is $a\sim b$ whenever $a-b$ is divisible by $24$. You can think of this as the set of hours in a day, because there's always a next hour, and always a previous hour, but we want to think of hours separated by $24$-hour increments as being "the same".

With a factor group (or, the more modern term "quotient group"), you have a group $G$ and a subgroup $H$. You want to know what happens when you force every element of $H$ to be equal to every other element, but still require the group operation to $\color{red}{(missing text)}$. (e.g. above, we set $0=24=48=\ldots$, which forced $1+0=1+24=1+48=\ldots$). We usually assume that $H$ is normal for technical reasons (if we glue together the elements of $H$, it turns out that we are forced to also glue together the elements of the smallest normal group containing $H$).

Then, with this little technical condition that $H$ is normal, we are just asking for $G/\sim$, where the equivalence relation is $x\sim y$ whenever $xy^{-1} \in H$. We glue together elements in such a way that everything in $H$ will be the identity of our new group. The even integers above are an example; another good one is the projection map $(x,y)\mapsto x$ from $\mathbb{R}^2\to\mathbb{R}$ as abelian groups—we are forcing every pair $(0,y)$ to be equal to $(0,0)$, which also forces $(x,y)$ to be equal to $(x,0)$ for every $x$. If you think about it, this is a demonstration that $\mathbb{R}^2 /\mathbb{R} \cong \mathbb{R}$ as abelian groups. More generally, we could describe the projection of a vector space $V$ onto a subspace $W$ as $V \to V/W^\perp \cong W$.

The quotient of a more complicated group is harder to visualize. But the basic ideas are all here. When you have a piece $H$ of a group $G$ that you want to ignore, the quotient group is almost always the way to do it. You are saying "This part ($H$) is unimportant, but let's see what happens when we get rid of it." The result of this thinking is $G/H$.

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What do you mean when you say we assume $H$ is normal for technical reasons? –  Jack M Feb 19 at 18:57
    
@JackM Essentially, if $H$ isn't a normal subgroup then just directly 'gluing together' elements that differ by a member of $H$ won't generate a clean subgroup. (Technical explanation: because the cosets of H will have non-trivial intersection; you can find elements $g_1, g_2\in G$ such that the sets $g_1H = \{g_1h: h\in H\}$ and $g_2H$ have some but not all elements in common.) –  Steven Stadnicki Feb 19 at 19:11

I suppose that $\mathbb Z_{60}/\langle 12 \rangle$ indicate the group obtained quotienting $\mathbb Z_{60}$ its subgroup generated by the element $12$: usually notation of the kind $\langle X \rangle$ means the smallest sub-structure (group, ring, module, ideal, etc) containing $X$, i.e. generated by $X$.

Generally when you have a group $G$ and a normal subgroup $H \lhd G$ the quotient group is the group having as support (or carrier) the set $$G/H = \{gH \mid g \in G\}$$ where each $gH$ is a coset and is a subset of $G$ defined as $$gH = \{ gh \mid h \in H\}$$

In case the group $G$ is abelian the is also used the notation $g+H$ to indicate the sets $gH$.

In the case you are considering the cosets $$x + \langle 12 \rangle = \{ x + 12*k \mid k \in \mathbb Z_{60}\}$$ more longly this set is the set $$\{x + 12, x + 24, x + 36, x + 48, x + 60 = x \}$$ which has cardinality $5$.

Since every element $x + \langle 12 \rangle$ is an element of the group $\mathbb Z_{60}/\langle 12 \rangle$ it makes sense to ask what is the order of such element in the group: i.e. for is the smallest positive integer $n \in \mathbb N \setminus \{0\}$ such that $\sum_{i=1}^n (x + \langle 12\rangle)=n(x+ \langle 12 \rangle) = 0$.

It's easy doing the count observe that $26 + \langle 12\rangle = 2 + \langle 12 \rangle$ and that the smallest $n \in \mathbb N \setminus \{0\}$ such that $$n(2 + \langle 12 \rangle) = 0 + \langle 12 \rangle = 12 + \langle 12 \rangle$$ is exactly $n=6$ which is the order of the element: indeed we have that such elements are $$1(2 + \langle 12 \rangle)=2 + \langle 12 \rangle$$ $$2(2 + \langle 12 \rangle)=4 + \langle 12 \rangle$$ $$3(2 + \langle 12 \rangle)=6 + \langle 12 \rangle$$ $$4(2 + \langle 12 \rangle)=8 + \langle 12 \rangle$$ $$5(2 + \langle 12 \rangle)=10 + \langle 12 \rangle$$ $$6(2 + \langle 12 \rangle)=12 + \langle 12 \rangle= \langle 12 \rangle$$ Hope this helps.

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All the cosets have the same cardinality, regardless of $x$. And the order is $6$, not $5$. –  Tobias Kildetoft Oct 8 '13 at 18:41
    
Although I agree with @TobiasKildetoft remark, this does really help clear it up. So when searching for order, I am interested in essentially the steps it takes to get to 0, which would be the steps from 0 to 12 in this case, and we're stepping up by 2. So, we have 2, 4, 6, 8, 10, 12 -> order of 6? –  David Oct 8 '13 at 18:47
    
@TobiasKildetoft yes sorry, thanks for pointing out. –  Giorgio Mossa Oct 8 '13 at 19:27
    
apparently hunger can help in making mistakes :P, it should be correct now. –  Giorgio Mossa Oct 8 '13 at 19:33
    
@David I have added the details of the proof that the order of $2+\langle 12 \rangle$ is $6$. –  Giorgio Mossa Oct 8 '13 at 19:37

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