Let $M$ be a maximum subgroup of $G,$ that is, a proper subgroup of $G$ such that the only subgroups of $G$ that contain $M$ are $M$ and $G.$ It shows that if $M\underline{\lhd}G$ then $[G:M]$ is finite and prime.
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Hi, could you explain a little bit about what you have tried so far? – Oliver Clarke May 14 '20 at 16:27
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Consider $x$ not in $M$. Clearly, the image of $x$ in $M$ is non-trivial. Note then that $\langle M,x\rangle$ strictly contains $M$, so that$\langle M,x \rangle=G$ and $G/M=\langle xM\rangle$. This means $G/M$ is a cyclic group generated by any of its non-trivial elements- i.e., it is isomorphic to the integers modulo p, for some prime p. Then we have that $[G:M]$ is finite and prime.
Aryaman Maithani
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TPace
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Because if y is not in M then it can be written as a product of elements of M and of x. – TPace May 14 '20 at 17:09