But what is with the other cyclic groups? Doesn't one also have to consider them?

92 Views Asked by At

I'm currently reading a textbook about abstract algebra.

There is a proof that every subgroup of a cyclic group is cyclic.

This proof is using the fact as every proof I have found on the Internet that all cyclic groups have the form $ \langle a\rangle=\{a^n\}$, where $n \in \mathbb{Z}$. But I don't think that this is true, because only cyclic groups under multiplication have this form.

But what is with the other cyclic groups? Doesn't one also have to consider them?

1

There are 1 best solutions below

0
On

It's just notation.

For each group $(G,\ast)$ (for any binary operation $\ast$ that defines a group on the set $G$), we may write the set $G$ under concatenation (which is the fancy term for putting symbols next to each other and it does not always denote multiplication), via the inclusion map $\iota(g)=g$ because $$\iota(g\ast h)=\iota(g)\iota(h)=gh$$ for arbitrary $g,h\in G$.${}^\dagger$ Powers, multiples, etc., depending on $\ast$, are then simply $g^n$ for arbitrary $g\in G, n\in\Bbb Z$.

This is to save time on writing/typing.

Concatenation may as well be the arbitrary notation to use for an arbitrary group, by fiat. In that case, we write $G$ instead of $(G,\ast)$ when the context is clear.


$\dagger$: Here $\iota$ is known as an isomorphism.