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?


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 $(1,2)$?

  • $\begingroup$ 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 $(a,b)\ne(0,0)$? $\endgroup$
    – Mike
    Mar 17 '12 at 22:06
  • $\begingroup$ @Mike, not quite. You need $(a,b)$ to belong to a $\mathbf{Z}$-module basis for that conclusion to hold. That happens, iff $\gcd(a,b)=1$. $\endgroup$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.