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Exercise from Herstein:Abstract Algebra.

Please do it without group action as I don't know it. In Herstein there is no mention of Group Action as in the answer given

Show that a group of order $108$ has a normal subgroup of order $9$ or $27$.

Attempt: $108=2^2\times 3^3$. If $n_2$ denotes the number of Sylow $2$ subgroups then $n_2=1+2k| 27\implies n_2=1,3,9,27.$

If $n_3$ denotes the number of Sylow $3$ subgroups then $n_3=1+3k| 8\implies n_3=1,2,4,8.$

If $n_3=1\implies $ the group has a normal subgroup of order $27$ but how to neglect the other choices.

I am confused totally .Please help.

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marked as duplicate by R_D, Dietrich Burde, Rohan, Namaste abstract-algebra Jan 10 '17 at 17:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ See here : math.niu.edu/~beachy/abstract_algebra/study_guide/… $\endgroup$ – Rohan Jan 10 '17 at 8:19
  • $\begingroup$ I don't know group actions@chítrungchâu $\endgroup$ – Learnmore Jan 10 '17 at 8:24
  • $\begingroup$ @chítrungchâu "order 9 or 27" is a stronger statement then "order $\ge$ 6" $\endgroup$ – Peter Jan 10 '17 at 8:27
  • $\begingroup$ just a way of expression, two posts have the same problem. $\endgroup$ – chí trung châu Jan 10 '17 at 8:30
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    $\begingroup$ You have written $n_3=1,2,4,8$, but you should have put $n_3=1$ or $4$. $\endgroup$ – Derek Holt Jan 10 '17 at 9:44
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$𝐺$ is a group of order $108=2^2⋅3^3$. The number of Sylow $3$-subgroups is either $1$ or $4$. Assume that it is $4$, otherwise we are done. Assume that $𝐻$ and $𝐾$ are two distinct Sylow $3$-subgroups of order $27$.

Let $𝑁=𝑁_𝐺(𝐻∩𝐾)$ be the normalizer of $𝐻∩𝐾$. We have to show that $𝑁=𝐺 $ i.e. $𝐻∩𝐾◃𝐺$.$$|𝐻∩𝐾|=\frac{|H||K|}{|HK|}≥\frac{|H||K|}{|G|}=\frac{27⋅27}{108}≈6.75.$$ Which forces $|H\cap K|$ to be $9$ so as to divide $|H|$ i.e. $27$.

We can also conclude that$𝐻∩𝐾◃𝐻$ and $𝐻∩𝐾◃𝐾$ since the index of $𝐻∩𝐾$ in each of $𝐻$ and $𝐾$ is $\frac{27}{9}=3$ and $3$ is the smallest prime divisor of $27$(i.e$ |𝐻|,|𝐾|$). Since $ 𝑁≤𝐺 $ is a subgroup we know $|𝑁|$ must divide $108=|𝐺|$. But $81=|𝐻𝐾|≤|𝑁|(why?)$ , so we must have $|𝑁|=108$ which implies $𝑁_𝐺(𝐻∩𝐾)=G$ i.e $(𝐻∩𝐾)$ is a normal subgroup of order $9$.

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