Can somebody explain me how is the converse of the Lagrange's theorem false by saying that $A_4$ (alternating group) doesn't contain any subgroup of order $6$?
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Do you know what "converse" means? If so, then your own post answers your own question! – Fly by Night Dec 01 '16 at 19:50
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Let $G=A_4$ and $G$ has a subgroup $H$ of order $6$. Since $|G:H|=2$, hence $H$ is normal in $G$. So, $G/H$ is a group of order $2$. Let $g\in G$ is an element of order $3$, i.e., $g^3=1$. We consider the coset $gH$. Since $G/H$ has order $2$, $(gH)^2=H$, i.e., $g^2H=H$, i.e., $g^{-1}H=H$ (as $g^2=g^{-1}$), i.e., $g\in H$. But $G=A_4$ has total $8$ elements of order $3$. Hence $|H|\geq 8$ which is a contradiction to the assumption of order of $H$. So, $A_4$ does not have any subgroup of order $6$. So this gives a proof of the fact that if $n$ divides order of a group $G$ then $G$ does not necessarily contain a subgroup of order $n$. Hence in general the converse of Lagrange's theorem is not true.
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Let N be the order of a group G. Then for that G, Lagrange's Theorem says: If a subgroup H of G exists, with H of order M, then M divides N. Then the converse is: If M divides N, then a subgroup H of G, with order M, exists.
Applied to A4, N = 12 and M = 6.
PMar
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I rephrase, how does a4 not have a subgroup of order 6. My bad, sorry. – Karan Bhasin Dec 01 '16 at 19:51
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Lagrange's Theorem says that if you have a group $G$ and a subgroup $H$ of $G$, then the order of $H$ must divide the order of $G$. The converse to this theorem is the following:
Suppose $G$ has order $n$. If $d$ divides $n$, then there is a subgroup $H$ of $G$ of order $d$.
Since the order of $A_{4}$ is $12$, if the above were true, it would follow that $A_{4}$ has a subgroup of order $6$, which it does not. Hence, the converse to Lagrange's statement is indeed false.
ervx
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