Finite group of even order has elements of order 2

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SUMMARY

A finite group G of even order contains at least one element of order 2. The proof utilizes the fact that if G has even order, the quotient group G/{1} has odd cardinality. By assuming no elements in G/{1} have order 2, it leads to a contradiction where G/{1} must have even cardinality. Consequently, there exists an element y in G/{1} such that y = y^-1, confirming the existence of an element of order 2.

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1MileCrash
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[The homework format does not appear on mobile]

Problem: Show that a finite group of even order has elements of order 2

Attempt:
The book gives a suggested approach that lead me to write the most round about, ugly proof I've ever written.

Can't I just say:
1.) If G has even order, G/{1} has odd cardinality.
2.) Assume that no elements of G/{1} has order 2.
3.) Then for each x in G/{1}, x^-1 is a distinct element of G/{1}.
4.) Then G/{1} has even cardinality.
Contradiction
5.) Therefore, there exists a y in G/{1} such that y = y^-1, and thus has order 2.
 
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1MileCrash said:
[The homework format does not appear on mobile]

Problem: Show that a finite group of even order has elements of order 2

Attempt:
The book gives a suggested approach that lead me to write the most round about, ugly proof I've ever written.

Can't I just say:
1.) If G has even order, G/{1} has odd cardinality.
2.) Assume that no elements of G/{1} has order 2.
3.) Then for each x in G/{1}, x^-1 is a distinct element of G/{1}.
4.) Then G/{1} has even cardinality.
Contradiction
5.) Therefore, there exists a y in G/{1} such that y = y^-1, and thus has order 2.

That is the standard method of showing that a finite group of even order has an element of order 2.
 

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