# Thread: cyclic groups

1. ## cyclic groups

Why is the set of integers under addition cyclic with generator 1? How can negative number be reached by continuosly adding 1?

2. You can check your definition of "cyclic group", but I'm pretty sure that when you claim a group is cyclic, you mean that every element of the group is (in the additive case) a multiple of one element or its inverse. So that's how you get your negative integers.

3. Originally Posted by Ackbeet
You can check your definition of "cyclic group", but I'm pretty sure that when you claim a group is cyclic, you mean that every element of the group is (in the additive case) a multiple of one element or its inverse. So that's how you get your negative integers.
Thanks for that. I wonder if you could help me out with the proof that subgroups of cyclic groups are themselves cyclic.

working; g=<a> , H is a subgroup of G
if H={e}, it is cyclic
if not, then a^n E H for some n

i am thinking i have to show a^n can generate every element of H .

4. It's been too long, my friend, since I took Abstract Algebra. Let me get another user to help you out here, one that knows algebra better than me!

Cheers.

5. Let G be a cyclic group, say $\displaystyle G = <g>$, and $\displaystyle H\leq G$. Since $\displaystyle H\leq G$, all elements of $\displaystyle H$ are of the form $\displaystyle g^k$ for some integer $\displaystyle k$. If $\displaystyle H = \{ e\}$, then $\displaystyle H$ is cyclic, so assume $\displaystyle \{ e\} < H$ and let $\displaystyle k$ be the least positive integer such that $\displaystyle g^k$ is in $\displaystyle H$. Let $\displaystyle g^s$ be any other element of $\displaystyle H$. By the division algorithm we can write $\displaystyle s = qk + r$ with $\displaystyle 0\leq r < k$. $\displaystyle g^r = g^{s-qk} = g^s(g^{-k})^q$ is in $\displaystyle H$ since $\displaystyle g^k$ and $\displaystyle g^s$ are. By the definition of $\displaystyle k$, $\displaystyle r$ must be $\displaystyle 0$. So $\displaystyle g^s = (g^k)^q$. and $\displaystyle H = <g^k>$.

6. Originally Posted by DrSteve
Let G be a cyclic group, say $\displaystyle G = <g>$, and $\displaystyle H\leq G$. Since $\displaystyle H\leq G$, all elements of $\displaystyle H$ are of the form $\displaystyle g^k$ for some integer $\displaystyle k$. If $\displaystyle H = \{ e\}$, then $\displaystyle H$ is cyclic, so assume $\displaystyle \{ e\} < H$ and let $\displaystyle k$ be the least positive integer such that $\displaystyle g^k$ is in $\displaystyle H$. Let $\displaystyle g^s$ be any other element of $\displaystyle H$. By the division algorithm we can write $\displaystyle s = qk + r$ with $\displaystyle 0\leq r < k$. $\displaystyle g^r = g^{s-qk} = g^s(g^{-k})^q$ is in $\displaystyle H$ since $\displaystyle g^k$ and $\displaystyle g^s$ are. By the definition of $\displaystyle k$, $\displaystyle r$ must be $\displaystyle 0$. So $\displaystyle g^s = (g^k)^q$. and $\displaystyle H = <g^k>$.
$\displaystyle (g^-k)^q$ isn't this a negative power?

7. I'm not sure what you mean by that. Yes, the exponent is negative. The important thing is that it's a memeber of H.