Finitely Generated Abelian Group

Just took my final exam and I wanted to see if I answered this correctly:

If $A$ is a Abelian group generated by $\left\{x,y,z\right\}$ and $\left\{x,y,z\right\}$ have the following relations:

$7x +5y +2z=0; \;\;\;\; 3x +3y =0; \;\;\;\; 13x +11y +2z=0$

does it follow that $A \cong Z_{3} \times Z_{3} \times Z_{6}$ ?

I know if we set $x=(1,0,2)$, $y=(0,1,0)$ and $z=(2,1,5)$ then this is consistent with the relations and with $A \cong Z_{3} \times Z_{3} \times Z_{6}$

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That is not correct, – i. m. soloveichik Dec 8 '12 at 20:33
With your $x,y,z$ I get $7x+5y+2z=(11,9,16)$; how is that $0$? – Gerry Myerson Dec 8 '12 at 21:54
Gerry you are correct. My apologies. I've made a correction for $z$ – Digital Gal Dec 9 '12 at 3:24
@jojo , it still is wrong: just behold the first coordinate: you get $\,9\,$ , not zero! – DonAntonio Dec 9 '12 at 3:33
9 mod 3 is 0 isn't it? – Digital Gal Dec 9 '12 at 3:36

Have you studied The Smith Normal Form of an (integer) square matrix? Well, if you form the matrix of coefficients of your relations you get:

$$A:=\begin{pmatrix}7&5&2\\3&3&0\\13&11&2\end{pmatrix}\Longrightarrow \det A=0$$

Thus, if $\,G:=\{x,y,z\}\,$ is the free abelian group on $\,\{x,y,z\}\,$ and $\,N\leq G\,$ is the (free abelian, of course) subgroup generated by the same letters but subject to the given relations, the quotient $\,G/N\,$ is finite iff $\,\operatorname{rank} N=3\Longleftrightarrow \det A\neq 0\,$.

From the above it follows your group cannot be what you wrote.

If you haven't studied the above then try to: it is very nice and spectacular stuff, though apparently you should reach the result otherwise.

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A colleague of mine has written some notes that we use in a course here. They should help you understand how to do this kind of question.

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Since $7x+5y+2z=0=3x+3y$, therefore $7x+5y+2z+2(3x+3y)=13x+11y+2z=0$, hence the last relation is not important. We also note that $7x+5y+2z-2(3x+3y)=x-y+2z$

Consider the group $G$={$ix+jy+kz|i,j,k\in Z$} (with addition defined as: ($i_1x+j_1y+k_1z)+(i_2x+j_2y+k_2z)=(i_1+i_2)x+(j_1+j_2)y+(k_1+k_2)z$).

Now let $N$ be the smallest subgroup of $G$ that contains $x-y+2z,3x+3y$. Thus, $N$={$i(x-y+2z)+j(3x+3y)|i,j\in Z$}={$(3j+i)x+(3j-i)y+2iz|i,j\in Z$}.

Thus, $A=G/N$. Now observe that $z$ has infinite order in $A$. Proof: Let $|z|=n>0$, therefore $nz=(3j+i)x+(3j-i)y+2iz$ for some $i,j$. This implies that $3j+i=3j-i=0$, hence $i,j=0$. Thus, $n=0$ (contradiction). Thus $A$ cant be isomorphic to the direct sum in your question

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Ooh! "definitely not $P$" is even stronger than "not necessarily $P$" (though only the latter was asked in the problem). – Hagen von Eitzen Dec 8 '12 at 22:39

The trivial counterexample is that the trivial group is generated by $x=0, y=0, z=0$ and of course the given relations hold with $x=y=z=0$. (Note that noone said that the set $\{x,y,z\}$ has cardinality $3$).

If you should insist on $x,y,z$ being distinct, observe that any quotient of $Z_3\times Z_3\times Z_6$ will preserve the relations. For example the projection to the first two factors $Z_3\times Z_3$ maps to the three distinct elements $(1,0), (0,1), (2,2)$.

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Isn't writing {x,y,z} a shorter way of saying that $x,y,x$ are distinct ? – Amr Dec 8 '12 at 20:37
@Amr. Not for me. $\{a,b\}$ is the set $S$ with $z\in S\iff z=a\lor z=b$. – Hagen von Eitzen Dec 8 '12 at 20:47
@Haegen von Eitzen OK I see – Amr Dec 8 '12 at 20:49
I think the group $A$ is the group {$x^{i}y^jz^k|i,j,k\in Z$}/$N$ where $N$ is the smallest subgroup containing $7x+5y+2z,3x+3y,13x+11y+2z$ – Amr Dec 8 '12 at 20:53
And the group operation is $x^{i_1}y^{j_1}z^{k_1}x^{i_2}y^{j_2}z^{k_2}=x^{i_1+i_2}y^{j_1+j_2}z^{k_1+k_2}$ – Amr Dec 8 '12 at 20:56