# Homework Help: Find possible subgroups given some elements

Tags:
1. Dec 6, 2016

### gummz

1. The problem statement, all variables and given/known data

Suppose that H is a subgroup of Z under addition and that H contains 2^50 and 3^50. What are the possibilities for H?

2. Relevant equations

Relevant concepts are just the definitions for a group and subgroup.

https://en.wikipedia.org/wiki/Group_(mathematics)

https://en.wikipedia.org/wiki/Subgroup

3. The attempt at a solution

The solution I was given is the following:

But what I'm wondering about, and would appreciate an answer to, is the following:

But the subgroup has operation "addition" so 2^50 = 50*2 = (2*5*5)*2, and 3^50 = 50*3 = (2*5*5)*3, so the possibilities for H are:

H=Z, H=2Z, H=10Z, H=25Z, H=50Z.

Or am I missing something?

2. Dec 6, 2016

### Staff: Mentor

Firstly, after $"H$ contains $\mathbb{Z}"$ you can stop, because $\mathbb{Z} \subseteq H \subseteq \mathbb{Z}$ (the latter for being a subgroup) already implies $H = \mathbb{Z}$.

What you are missing is, that $2^{50} \neq 50^2$ or $100$.

3. Dec 6, 2016

### gummz

Thanks for reply. But if we write out the notation, then, for group with operation "addition":

250 = 50*2, and 502=2*50, and Z is commutative? That is, in general, an=n*a

4. Dec 6, 2016

### Staff: Mentor

No, you can't confuse these notations. $2^{50}$ is simply a number and the subgroup is $(H,+)=2^{50}\cdot (\mathbb{Z},+) + 3^{50}\cdot (\mathbb{Z},+)$. Otherwise one would have defined $H=100\mathbb{Z}+150\mathbb{Z}=50\mathbb{Z}$.

5. Dec 7, 2016

### Stephen Tashi

I think the notation used in the problem isn't consistent with the interpretation of $a^k$ for a group element $a$. You are correct that for a group operation denoted "$*$", the interpretation of the notation $a^3$ is usually $a*a*a$. So for the operation "$+$", the interpretation would be "$a+a+a$". However, I think the problem is using the notation for exponents in the standard sense that it is used for the field of real numbers. By that notation $2^3 = (2)(2)(2)$ instead of $2+2+2$.