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I've somewhat recently been going back through one of my brother's old textbooks reviewing group theory. I'm up to a chapter called Factor-Group Computations and Simple Groups. The problems at the end seem to have me stumped and I want to make sure I'm understanding enough before I proceed.

There are a dozen problems asking to classify a given group according to the fundamental theorem of finitely generated abelian groups. These are a couple that stumped me.

The first is $(Z_4\times Z_4\times Z_8)/\langle (1,2,4)\rangle$. I can see that $Z_4\times Z_4\times Z_8$ has order $128$ while $\langle(1,2,4)\rangle$ has order $4$, so the factor group in question has order $32$. It also appears there are elements of up to order $8$, which I think should narrow it down to $Z_4\times Z_8$ or $Z_2\times Z_2\times Z_8$, but how do I tell which?

This next one I have no clue how to proceed with. $(Z\times Z)/\langle(1,2)\rangle$ How would I go about solving this one?

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2 Answers 2

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A hint for your first question: Subtracting a multiple of $(1,2,4)$ from any element $(a,b,c)$ of $\mathbf{Z}_4\times\mathbf{Z}_4\times\mathbf{Z}_8$ allows us to make the first coordinate equal to zero. So within each coset of $\langle(1,2,4)\rangle$ there is an element with first coordinate equal to zero. Why does this imply that the quotient group can be generated by at most two elements?

A hint for your second question: Can you think of a basis of $\mathbf{Z}\times\mathbf{Z}$ containing the element $(1,2)$?

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All right, I think I see where the first argument is going, though maybe I should work it out for myself later just to make sure. As for the second question, does this mean $(Z\times Z)/<(a,b)>$ is isomorphic to $Z$ whenever $(a,b)\ne(0,0)$? –  Mike Mar 17 '12 at 22:06
    
@Mike, not quite. You need $(a,b)$ to belong to a $\mathbf{Z}$-module basis for that conclusion to hold. That happens, iff $\gcd(a,b)=1$. –  Jyrki Lahtonen Mar 18 '12 at 5:52

Method 1. One can always use the Smith Normal Form when trying to determine the isomorphism type of subgroups and quotients of $\mathbb{Z}^r$.

So we can apply it directly to the second problem. As for the first, we can use the isomorphism theorems: remember that we can view $\mathbf{Z}_4\times\mathbf{Z}_4\times\mathbf{Z}_8$ as a quotient of $\mathbf{Z}\times\mathbf{Z}\times\mathbf{Z}$ by the subgroup $\langle (4,0,0), (0,4,0), (0,0,8)\rangle$. The subgroup generated by $(1,2,4)$ in the quotient corresponds to the subgroup $$\langle (4,0,0), (0,4,0), (0,0,8), (1,2,4)\rangle$$ of $\mathbf{Z}\times\mathbf{Z}\times\mathbf{Z}$. Using the Smith Normal Form, we can then obtain a "nice" basis for the latter group and the subgroup in question (where each element of the basis of the subgroup is an integer multiple of a distinct element of the basis of the overgroup), and easily determine the isomorphism type of the quotient.


Method 2. For a more pedestrian approach to the first problem:

Remember that if $A$ is an abelian group, then a subset $a_1,\ldots,a_n$ is "independent" if and only if whenever $m_1a_1+\cdots+m_na_n=0$, with $m_i\in\mathbb{Z}$, then $m_ia_i=0$ for all $i$; this is a generalization of the usual notion of "linearly independent".

I claim that $(1,2,4)$, $(0,1,0)$, and $(0,0,1)$ are independent in $A=\mathbf{Z}_4\times\mathbf{Z}_4\times\mathbf{Z}_8$, and that they generate the group.

The fact that the generate is easy: we can obtain $(1,0,0)$ from these three as $$(1,0,0) = (1,2,4) - 2(0,1,0) - 4(0,0,1),$$ hence $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ are all in the subgroup generated by the three elements; since these plainly generate $A$, our set does as well.

As for independent, suppose that $$m_1(1,2,4) + m_2(0,1,0) + m_3(0,0,1) = (0,0,0).$$ Then $m_1\equiv 0\pmod{4}$, $2m_1+m_2\equiv 0\pmod{4}$, and $4m_1+m_3\equiv 0\pmod{8}$. From these, it easily follows that $4|m_1$, $4|m_2$, and $8|m_3$, so $m_1(1,2,4)=m_2(0,1,0)=m_3(0,0,1) = (0,0,0)$.

This means that we can express $A$ as $$A = \langle (1,2,4)\rangle \times \langle (0,1,0)\rangle \times \langle (0,0,1)\rangle.$$ Figuring out what happens when you mod out by $\langle(1,2,4)\rangle$ is now trivial.

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