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boboYO

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'Prove that if a finite group G has only one maximal subgroup M, then |G| is the power of a prime'

I've somehow deduced that no finite group has only one maximal subgroup, and I'm having trouble seeing where I went wrong.

This is what I have:

Let [tex]H_1[/tex] be a subgroup of G. Either [tex]H_1[/tex] is maximal and equal to M or it is not maximal and there is a [tex]H_2[/tex] such that [tex]H_1<H_2<G[/tex] (using < to mean proper subgroup). Apply the same argument to [tex]H_2[/tex] and we get an ascending chain of subgroups. Since G is finite the process must end eventually (when we reach M).

Thus every subgroup of G is a subgroup of M and hence every element of G is an element of M, and G=M, a contradiction.

So would someone like to point out the flaw in the above reasoning? thanks.

I've somehow deduced that no finite group has only one maximal subgroup, and I'm having trouble seeing where I went wrong.

This is what I have:

Let [tex]H_1[/tex] be a subgroup of G. Either [tex]H_1[/tex] is maximal and equal to M or it is not maximal and there is a [tex]H_2[/tex] such that [tex]H_1<H_2<G[/tex] (using < to mean proper subgroup). Apply the same argument to [tex]H_2[/tex] and we get an ascending chain of subgroups. Since G is finite the process must end eventually (when we reach M).

Thus every subgroup of G is a subgroup of M and hence every element of G is an element of M, and G=M, a contradiction.

So would someone like to point out the flaw in the above reasoning? thanks.

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