0

Find all subgroups of $\mathbb{Z_2} \times \mathbb{Z_5} \times \mathbb{Z_7}$.

I know that this would be isomorphic to $\mathbb{Z_{70}}$, so that I would be looking for $8$ subgroups because $70$ has $8$ factors.

I tried finding an isomorphism between $\mathbb{Z_2} \times \mathbb{Z_5} \times \mathbb{Z_7}$ and $\mathbb{Z_{70}}$, but it seems like an unreasonable about of tedious calculation to find some way to represent all the elements of the subgroups of $\mathbb{Z_{70}}$ in the form $ax+by+cz$, where the corresponding element in $\mathbb{Z_2} \times \mathbb{Z_5} \times \mathbb{Z_7}$ would be ${(x,y,z)}$. Am I just thinking lazy and need to bash this one out? Or am I approaching the problem in the wrong way? Any help would be appreciated.

MrMustache
  • 83
  • This group is isomorphic to $\Bbb Z/70\Bbb Z$, as you wrote. The subgroups are in bijection to the groups $d\Bbb Z/70\Bbb Z$, where $d$ is a positive divisor of $70$. – Dietrich Burde Feb 12 '24 at 17:23
  • If you aren't so lucky as to have a cyclic group, Goursat's lemma tells you how to find the subgroups of a product. – While I Am Feb 12 '24 at 17:25
  • @William Yes, indeed. This can be found here, for example. But fortunately, as you said, this here is much easier, and has been answered several times already. – Dietrich Burde Feb 12 '24 at 17:26

1 Answers1

2

Subgroups must be of size 1,2,5,7,10,14,35,70 by Lagrange's theorem:

${e}$ :size 1

${Z_{2}\times Z_{5} \times Z_7}$ :size 70

Now consider size 35: We must not be able to generate only half the elements of the group.

$(e,x^i,y^j)$ does this. And no other way to do that , as the groups in the product have prime order importantly, so the order of those groups are prime, so each element generates every element (apart from the identity which does not obviously)

Size 14: $(a^m,e,y^j)$ is of size 14 . And again, there is no other way to do this, as those other groups have prime order , so all elements are cyclic apart from the identity.

Size 10:

$(a^m,y^j,e)$...

Size 5:

$(e,y^j,e)$...

You get the idea. I leave the rest to complete yourself. The key idea is that because the product of the groups uses groups of prime order, and so there is only one such possible subgroup configuration for each size allowed by lagrange's theorem.

J.Dmaths
  • 715