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Gallian says in Chapter 3 of Contemporary Abstract Algebra that

For any element $a$ of a group $G$, it is useful to think of $\langle a \rangle$ as the smallest subgroup of $G$ containing $a$.

But wouldn't this mean the set $\langle a \rangle$ is simply $\{ a, a^{-1}, e \}$?

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    $\begingroup$ If it doesn't contain $a^2$, that isn't a subgroup. $\endgroup$ – Robert Israel Apr 10 '17 at 16:00
  • $\begingroup$ @RobertIsrael Why not? $\endgroup$ – Airdish Apr 10 '17 at 16:03
  • $\begingroup$ It needs to be closed under the group's operation. $\endgroup$ – Akiva Weinberger Apr 10 '17 at 16:58
  • $\begingroup$ That isn't a group unless $a = a^{-1}$, in which case it would be $\{a, e\}$ only $\endgroup$ – q.Then Apr 10 '17 at 17:53
  • $\begingroup$ @q.Then Actually that is a group if $a^2 = a^{-1}$. $\endgroup$ – fleablood Apr 10 '17 at 18:09
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The set $\{a,a^{-1},e\}$ has an identity and is closed under inverses, but it isn't necessarily closed under the group's operation.

If $H$ is a subgroup and $a \in H$, you also need $a^2=aa\in H$ and $a^3=aaa \in H$ etc. Likewise, you need inverses for all of those elements (i.e. $a^{-1}, a^{-2}, \dots \in H$).

So closure under the group's operation and inverses require at least $\langle a \rangle = \{ a^n \;|\; n \in \mathbb{Z} \} \subseteq H$.

Once you know $\langle a \rangle$ is itself a subgroup. This shows that it is the smallest subgroup containing $a$.

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    $\begingroup$ So what you're saying is that once you have $a$ in the subgroup, you automatically need its powers too because $a * a$ and $a *a * a$, etc. ought to be allowed operations under the group operation? $\endgroup$ – Airdish Apr 10 '17 at 16:05
  • $\begingroup$ A group's operation is associative, has an identity, and inverses...but it also must be closed. If $a,b \in G$, then $ab \in G$ as well. So this means $a \in G$ then $a^2=aa \in G$ etc. So closure is what is forcing this. $\endgroup$ – Bill Cook Apr 10 '17 at 16:06
  • $\begingroup$ Ah okay so the set $\{ a, a^{-1}, e \}$ isn't actually a group at all? $\endgroup$ – Airdish Apr 10 '17 at 16:08
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    $\begingroup$ Not in general. For example: let $i=\sqrt{-1}$. Then $\{i,i^{-1},e\} = \{i,-i,1\}$ is not a group since multiplication isn't a closed binary operation operation on this set ($i^2=-1 \not\in \{i,-i,1\}$. Since we don't have a closed operation, it's not a group. $\endgroup$ – Bill Cook Apr 10 '17 at 16:12
  • $\begingroup$ On the other hand, it could be. Consider $a=-1$ (with multiplication). Then $\{a,a^{-1},e\}=\{-1,(-1)^{-1},1\} = \{1,-1\}$ which is a perfectly good (cyclic) group (under multiplication). So $\{a,a^{-1},e\}$ can be a group, but usually it isn't. :) $\endgroup$ – Bill Cook Apr 10 '17 at 16:14
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Groups are closed under their operations. In other words if $a,b \in G$ then we know that $ab \in G$.

So if $a\in <a>$ then we know that $a*a=a^2 \in <a>$ and $a^2*a = a^3 \in <a>$ and inductively we know that $a^2 \in <a>$ for all $n \in \mathbb N$. And as inverses must exist we know $a^{-n} \in G$.

So in general the group $<a> = \{.......a^{-3},a^{-2},a^{-1}, e,a,a^2,a^3,....\}$. But the $a^i$ need not be distinct. It could be possible for $a^i = a^k;i \ne j$. An example is $\mathbb Z_6=\{0,1,2,3,4,5\}$ under modulo addition. $2^5 = 2+2+2+2+2 = 4 = 2+2 = 2^2$. So $\{......2^{-3}=0,2^{-2}=2, 2^{-1}=4, 0, 2, 2^2 = 4, 2^3 = 0,....\} = \{0,2,4\}$ is not infinite.

Now it is possible that $a^k =e$ for some number $k$. Then $a^{-1} = a^{k-1}$ Example: In $\mathbb Z_6$, $2^3 = 2+2+2= 0$ and $2^{-1} = 4 = 2^{3-1}$. In this case $<a> = \{0, a, a^2, a^3, .... a^{k-1}\}$. Example $<2> \subset \mathbb Z_{6} = \{0,2,2^2=2^{-1}=4\}$

But it's also possible that $a^k \ne e$ for any $k \ne 0$. In this case $<a> = \{a^n|n \in \mathbb Z\}$. Example: If the group is $\mathbb Z$ under addition. Then $<7> = \{0, 7, -7, 14, -14, 21, -21, ....\} = \{7n| n \in \mathbb Z\}$. Which is infinite.

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In general. If $a^n= e$ and $a^k\ne e; 1\le k < n$ we so "the order of $a$" is $n$ and we write it as $|a| = n$. If no such $n$ exists, we say $|a| = \infty$.

If $|a| = n$ then $<a> =\{0,a, a^2, ....., a^{n-1}\}$.

If $|a| = \infty$ then $<a> = \{a^k|k \in \mathbb Z\}$.

Notice $<a>$ always has $|a|$ units.

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