# Cyclic Groups proof help

## Homework Statement

Let G be a group. Assume a to be an element of the group. Then the set <a> = {ak I k∈ℤ} is a subgroup of G.

I am confused as to why the proof makes the assumption that <a> is a subset of the set G.

## The Attempt at a Solution

The proof I think is like the following:
As the identity element is in G it is true that it is also in <a>. Since for a general group G, the inverse is denoted as a0. Let b=ar and c=aj be elements in the group then araj is an element in the group due to the axiom of exponents... Now proving that there exists an inverse is done in an similar way but even though these three conditions are satisfied, shouldn't I prove that <a> is a subset of G?[/B]

Stephen Tashi
I am confused as to why the proof makes the assumption that <a> is a subset of the set G.

What proof are you talking about? Is it a proof that you read?

Or are you asking whether a proof that you yourself write should prove <a> is a subset of G?

shouldn't I prove that <a> is a subset of G?

Yes you should.

If we split hairs, you should show that the phrase "the set ##<a> = \{ a^k | k \in \mathbb{Z}\}##" actually defines a set. After all, we can write down phrases that don't define specific sets - such as "##W = \{a+k | k \in \mathbb{Z}\}##".

You could use induction to show that ##a^k## has a defined result that is an element of ##G##.

Your approach beginning "Let ##b=a^r## and ##c=a^j## be elements in the group" only shows that if ##a^r## and ##b^j## are elements of ##G## then ##a^{r+j} ## is an element of ##G##. You hypothesize that ##a^r## is an element of ##G## instead of proving it and you didn't mention that ##a^{r+j}## is an element of ##<a>##.

member 587159
Notice that a group operation is a function ##\circ: G \times G \to G##. That is, the product of 2 groups element is again in the group. Can you see how to apply that to find that ##a^k \in G## for ##a \in G, k \in \mathbb{Z}##?

Delta2
fresh_42
Mentor
shouldn't I prove that <a> is a subset of G?
Formally, yes. But as its proof is more or less straight forward, some authors might not feel the need to actually do it.