# Show that a group of order $108$ has a normal subgroup of order $9$ or $27$. [duplicate]

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.

## marked as duplicate by R_D, Dietrich Burde, Rohan, Namaste abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 10 '17 at 17:39

• – Rohan Jan 10 '17 at 8:19
• I don't know group actions@chítrungchâu – Learnmore Jan 10 '17 at 8:24
• @chítrungchâu "order 9 or 27" is a stronger statement then "order $\ge$ 6" – Peter Jan 10 '17 at 8:27
• just a way of expression, two posts have the same problem. – chí trung châu Jan 10 '17 at 8:30
• You have written $n_3=1,2,4,8$, but you should have put $n_3=1$ or $4$. – Derek Holt Jan 10 '17 at 9:44

$$𝐺$$ 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 $$9$$.