"Prove that if G is a finite group and H is a proper normal subgroup of largest order, then G/H is simple."

Question

Prove that if G is a finite group and H is a proper normal subgroup of largest order, then G/H is simple.

how to reach G/H . ITS FROM GALLIAN ?

This question is actually only the backward step of a much more useful fact, that $G/H$ is simple $\iff$ $H \lhd G$ is maximal. All of the answers in this SE thread will satisfy your question: http://math.stackexchange.com/questions/161570/h-is-a-maximal-normal-subgroup-of-g-if-and-only-if-g-h-is-simple — user113525, May 14 '14 at 18:06

score 1 · Answer 1 · answered May 14 '14 at 18:18

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Hint: The inverse image of a normal subgroup of $G/H$ is a normal subgroup of $G$ that contains $H$. Try to argue that this inverse image has to be $G$ or $H$.

answered May 14 '14 at 18:18

mez

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score 0 · Accepted Answer · answered May 14 '14 at 18:06

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If $H$ is maximal wrt normal and order, then one cannot find a normal subgroup $K$ with $H \subsetneq K \subsetneq G$. Hence $G/H$ is simple.

answered May 14 '14 at 18:06

Nicky Hekster

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You intend strict inclusions I think – Geoff Robinson May 14 '14 at 18:23
Geoff, absolutely, contradicting maximality. – Nicky Hekster May 14 '14 at 18:25
Yes, your intention was clear, but your notation allows $K =H$ or $K=G. $ – Geoff Robinson May 14 '14 at 19:00

"Prove that if G is a finite group and H is a proper normal subgroup of largest order, then G/H is simple."

2 Answers2