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Consider $26 + <12>$ in $\mathbb{Z}_{60} / <12>$, I need to find the order of the element in this quotient group.

So I basically wrote out what the group might look like

$\dfrac{\mathbb{Z_{60}}}{<12>} =\{<12>,1+<12>, 2+<12>,3+<12> , \dots\ 11 + <12> \}$

So $26 + <12> = \{26,38, 50,2,14 \}$, so $2$ is in the coset and therefore $26+<12> = 2+<12>$.

So to ask for the element of the quotient group is the same as asking the order of cyclic subgroup of the quotient group. So I need to find all $n(2+<12>)$ for some integer $n$

So for instance $n=0$, I get $0(2 + <12>) = 0 + <12> = <12>$.

Now aftering trying out $n = 6$, I get back $<12>$,does this show that $<12>$ is identity? Because I tried the definition of identity $x \star e = x$ and $2 + <12> + e = 2 + <12> \iff e = 2 + <12> - (2 + <12>) = 2 + <12> - 2 - <12> = 0$?

Does the negative sign distribute over to the $<12>$?

Questions

  1. Does the negative sign distribute over to the $<12>$?

  2. My book just skips all the details. I am not even sure how this is obvious, but how do they immediately know that $26 + <12> = 2 + <12>$ without even writing out $26 + <12>$ first?

3.When I wrote that

$n=0$, I get $0(2 + <12>) = 0 + <12> = <12>$.

Is this even right? Can I even write out like this? Why doesn't $0$ distribute over?

  1. Please give me your harshest criticism of how I can write this out better.
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Step 1: you should write \langle and \rangle instead of < and >. –  Alexander Gruber Nov 29 '12 at 14:27
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1 Answer

up vote 2 down vote accepted

The <26> is there to symbolize that your element is a coset in the quotient group, i.e. that you are meaning $ a + <12> = \{a' \in \mathbb{Z} : a' - a \in <12> \}$ or, in other terms, $a' - a$ is a multiple of 12.

So, to answer the first question, no, the minus sign does not distribute to the <12>. What you're doing when you add (or subtract) is taking the sum (or difference) of the cosets. And since the quotient group $\mathbb{Z}_{60}/<12>$ is another group, the sum of two elements (cosets) will be again another coset.

For the second question, why does 26 + <12> = 2 + <12>? If you take out the "12 part of 26", i.e., write $26 = 2 + 12 + 12$, then the coset of 26 modulo 12 will be the set of all elements $\{z \in \mathbb{Z} : z-26 = 12 k \}$, and you may say that 2 satisfies this, so 2 and 26 behave exactly the same.

In essence, by taking the quotient, we are mapping 12 to zero, and seeing what happens to the group structure.

Third question: Since you're looking at this structure as a group under addition, you don't really have multiplication defined, unless by $0(2 + <12>)$, you mean iterated addition (adding 2 + <12> to itself 0 times) which is defined to be the identity, 0 + <12> in the group.

In response to your request for criticism, I think you need to find a good book and read and love algebra. Quotient groups are a difficult concept to wrap your head around, but once you'll get it you'll develop intuition for it and see the structure and beauty of it.

I hope this helps. A good reference is Dummit and Foote's Abstract Algebra, and Artin's Algebra.

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