Unifying Orders in Finite Group Conjugacy Classes

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In a finite group, all elements within a conjugacy class share the same order. The conjugacy class cl(a) is defined as {xax^-1 | x in G}, where G is a finite group. It is established that the identity element e and all elements ci in the conjugacy class cl(g) have orders that divide n, the size of the conjugacy class. The discussion emphasizes understanding the relationship between the order of an element and the exponent n, particularly noting that if x^n = e, it implies a connection between n and the order of e. Ultimately, this leads to the conclusion that elements in a conjugacy class must have the same order.
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Show that the
elements in a conujgacy class of a
finite group all have the same order

cl(a) = {xax^-1|x in G} G is finite
G = {e,g1,g2,...,gm}

cl(g)={e,c1,c2,c3,...,cn} finite for n =< m

Then |e| | n , |c1| | n, |c2| | n, |cn| | n


Well C1 =xgix^1 for some gi.

any hint?
 
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What is the definition of order of an element? And I don't mean for you to just post it here, I mean for you to go from that and think about things: if x^n=e what can you say about n relative to the order of e?
(That the conjugacy class sizes divide the order of the group is immaterial.)
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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