The following notation is used in my teacher's notes for an abstract algebra course. It states that let G be a cylic group such that $G = $ { $e,g^{\pm 1}, g^{\pm2}, ... $}. Why is each exponent written in $\pm$ ? I have seen other instances where $G$ is described as $G = <g> = $ { $e, g^1, g^2... $}. Hopefully this question is clear thanks.
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1Maybe it's explicitly showing that each element has an inverse? – Edward Evans Mar 21 '16 at 23:48
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OK that makes sense. – IntegrateThis Mar 21 '16 at 23:49
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3${e, g^1, g^2, \ldots}$ is not a group unless $G$ is finite, in which case $g^N = e$ for some positive $N$, and then inverses are included in ${e, g^1, g^2, \ldots, g^{N-1}}$ because $g^{-k} = g^{N-k}$. – Mar 21 '16 at 23:51
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Usually, the subgroup generated by some elements $\{g_1,g_2,\cdots\}$ denoted by $$<g_1,g_2,\cdots>$$ is defined as the subgroup containig all those elements, their inverses, and the products you can make with them. This is equivalent to the smallest subgroup containing $g_1,g_2,\cdots$
In general
If we would leave out the part about the inverses, this wouldn't be equivalent. For example in the group $(\mathbb{Z},+,0)$, the set $\{3\}$ would generate $$\{3, 6, 9, \cdots\}$$ which is not a subgroup. The part about the inverses deals with that problem and makes it $$\{\cdots,-6,-3,0,3,6,\cdots\}\cong \mathbb{Z}$$ Which is indeed the smallest subgroup containing $3$.
Your professor just wrote down this explicit definition. In the finite case, this is not required but the above example shows it is required in general. The other places were probably talking about finite groups
Finite groups
The result we got when leaving out the part about the inverses is clearly the smallest monoid containing these elements. Since it is a submonoid of a group, the cancelation law holds. Since it's finite, it's guaranteed to be a group
Jens Renders
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