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Suppose [itex] a \in <b>[/itex]

Then [itex]<a> = <b>[/itex] iff a and b have the same order (let the order be n - the group is assumed to be finite for the problem).

Proof:

Suppose a and b have the same order (going this direction I'm trying to show that <a> is contained in <b> and <b> is contained in <a> ergo <a> = <b>).

Since [itex]a \in <b>[/itex] it is obvious that [itex]<a> \subseteq <b>[/itex].

Since a is in <b>, [itex]a = b^{m}[/itex] for some m.

So [itex]<a> = <b^{m}> \supseteq <b>[/itex]

Hence <a> = <b>

Suppose <a> = <b> and <b> has order n.

[itex]a = b^{m}[/itex]'

[itex]a^{n} = (b^{m})^{n} = (b^{n})^{m} = e^{m} = e[/itex]

Hence a has order n.

This all look good?

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# Cyclic subgroup proof

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