# Groups. I am stumped again.

1. May 17, 2004

### Nexus[Free-DC]

Okay, I'm really scratching my head here.

If an Abelian group A has three generators x,y,z and they are subject to three defining relations, say something like

x+y+z=0
x-y-z=0
2x-2y+3z=0

then I can solve for x,y,z and find A as a direct sum of cyclic groups, Z_x + Z_y + Z_z.

But what do I do if the three equations are not linearly independent? I get left with everything in terms of x and I can't just plug in the numbers.

Thanks,
N.

2. May 17, 2004

### NateTG

I don't understand

There's some context missing:

Let's say you've got generators $$\{x,y,z\}$$ and the relations:
$$xy=yx$$
$$yz=zy$$
and
$$zx=xz$$
Which are equivalent to the claim that the group is abelian.

Now we add the following relations:
$$0=0$$

So at this point, this group is equivalent to $$\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}$$ and the system of defining relations is clearly not linearly independant.

3. May 17, 2004

### matt grime

you probably oughtn't to confuse additive and multiplicative notation in the same thread

4. Sep 17, 2004

### mathwonk

nexus, you are talking about a representation of your abelian group as a cokernel of a map which is not necessarily injective.

by the way your example is non sense as your groupo as described is zero.

i.e. you gave so many relations that everything was trivial.

In general if you have n generators, that emans you map a direct sum of n copies of Z onto the group. then telling what say r relations are, is giving a generating set for the kernel of that map, hence it lets you map another direct sum of r copies of Z onto the kernel.

So now you have a linear map from r copies of Z, to n copies of Z, given by a matrix of integers. and your group is the cokernel of this map.

To get the explicit structure of the group, you just diagonalize the matrix using row and column operations. Then you can easily see the group structure, as the quotient of the direct sums then becoems the direct sum of the quotients of the diagonal maps, i.e. one dimensional maps.

this is the usual proof of the structure theorem for finitely generated abelian groups.