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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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    See here : http://www.math.niu.edu/~beachy/abstract_algebra/study_guide/soln74.html#s7418 – 2017-01-10
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    I don't know group actions@chítrungchâu – 2017-01-10
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    @chítrungchâu "order 9 or 27" is a stronger statement then "order $\ge$ 6" – 2017-01-10
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    just a way of expression, two posts have the same problem. – 2017-01-10
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    $BenStokes I think you have to know group action or Sylow theorem to solve this – 2017-01-10
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    You have written $n_3=1,2,4,8$, but you should have put $n_3=1$ or $4$. – 2017-01-10

1 Answers 1

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$$ is a group of order $108=2^2⋅3^3$. The number of Sylow $3$-subgroups is either $1$ or $4$. Assume that it is $4$, otherwise we are done. Assume that $$ and $$ are two distinct Sylow $3$-subgroups of order $27$.

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

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