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Martin Sleziak
Martin Sleziak
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Seen on http://legauss.blogspot.com.es/2012/05/para-rir-ou-para-chorar-parte-13.htmlhttps://legauss.blogspot.com/2012/05/para-rir-ou-para-chorar-parte-13.html

Theorem: $5!/2$ is even.

Proof: $5!/2$ is the order of the group $A_5$. It is known that $A_5$ is a non-abelian simple group. Therefore $A_5$ is not solvable. But the Feit-Thompson Theorem asserts that every finite group with odd cardinal is solvable, so $5!/2$ must be an even number.

Seen on http://legauss.blogspot.com.es/2012/05/para-rir-ou-para-chorar-parte-13.html

Theorem: $5!/2$ is even.

Proof: $5!/2$ is the order of the group $A_5$. It is known that $A_5$ is a non-abelian simple group. Therefore $A_5$ is not solvable. But the Feit-Thompson Theorem asserts that every finite group with odd cardinal is solvable, so $5!/2$ must be an even number.

Seen on https://legauss.blogspot.com/2012/05/para-rir-ou-para-chorar-parte-13.html

Theorem: $5!/2$ is even.

Proof: $5!/2$ is the order of the group $A_5$. It is known that $A_5$ is a non-abelian simple group. Therefore $A_5$ is not solvable. But the Feit-Thompson Theorem asserts that every finite group with odd cardinal is solvable, so $5!/2$ must be an even number.

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Ramón Barral
Ramón Barral
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Seen on http://legauss.blogspot.com.es/2012/05/para-rir-ou-para-chorar-parte-13.html

Theorem: $5!/2$ is even.

Proof: $5!/2$ is the order of the group $A_5$. It is known that $A_5$ is a non-abelian simple group. Therefore $A_5$ is not solvable. But the Feit-Thompson Theorem asserts that every finite group with odd cardinal is solvable, so $5!/2$ must be an even number.