I'm finding this question particularly difficult to answer. Could someone help me, please?

Edit: I have seen the use of the notation ker which I know translates to kernel, in some solutions, but this notation is not used in my course on Group Theory. What is the alternative solution to this?

Armstrong question 11.11

  • $\begingroup$ Hint: Consider the cosets $H$, $Hg$, $Hg^2$, ... $\endgroup$ – Derek Holt Jun 2 '17 at 11:32
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    $\begingroup$ See also math.stackexchange.com/questions/545417/… $\endgroup$ – lhf Jun 2 '17 at 12:07
  • $\begingroup$ But I wonder what the book intends by suggesting that we "adapt the proof of Lagrange's Theorem". $\endgroup$ – lhf Jun 2 '17 at 13:16

Let $X$ be the set of left cosets of $H$. Then $\phi: G \to \text{Sym}(X)$ given by $\phi(g)(aH)=(ga)H$ is a homomorphism. Therefore, by Lagrange's Theorem, $id = \phi(g)^{m!}= \phi(g^{m!})$ for all $g \in G$. In particular, $H=\phi(g^{m!})(H)=g^{m!}H$ and so $g^{m!} \in H$.

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  • $\begingroup$ I'm not sure this answer fits the book's suggestion that we "adapt the proof of Lagrange's Theorem". $\endgroup$ – lhf Jun 2 '17 at 13:20
  • $\begingroup$ Yes, that must be what is meant by that condition. $\endgroup$ – arctic tern Jun 3 '17 at 0:48
  • $\begingroup$ I know this is a novice question to ask, but what is the definition of the notation Sym? That notation has never appeared in my textbook, so I don't really understand what is going on here. $\endgroup$ – wsh_97 Jun 3 '17 at 0:48
  • $\begingroup$ @Goooie, $\text{Sym}(X)$ is the group of bijections (permutations) of $X$. $\endgroup$ – lhf Jun 3 '17 at 1:59

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