If the order of a non abelian group $G$ is $39$, then find all normal subgroup in $G$.

Question

If the order of a non abelian group $G$ is $39$, then find all normal subgroup in $G$.

All I know is a subgroup $N$ is normal in group $G$:

(1) if $N$ is abelian or cyclic in $G$.

(2) if $N$'s index is $2$ in the group $G$.

(3) if $N$ is either trivial or the whole $G$.

Also if order of a group $G$ is $pq$ ($p, q$ prime and $p<q$) then there exist unique subgroup of order $q$ and hence is normal (since it is isomorphic to $\Bbb Z_q$) and also total subgroup of order $p$ is equal to $q.$

Answer 1

Trivially, $1$ and $G$ are normal subgroups.

From the Sylow theorems, it follows that there is exactly one subgroup of order $13$; this is therefore characteristic and hence normal. So far we have three normal subgroups.

The remaining $39-13$ elements of $G$ must be of order $3$ (because any element of order $39$ would make $G$ cyclic, hence abelian). In particular, there are $13$ Sylow $3$-groups. None of these is normal (as $G$ acts transitively on the set of Sylow $3$-groups by conjugation).

Answer 2

Order of the group $G$ is 39 i.e. $\circ (G)=3.13$. Now since $3\mid (13-1)$. Therefore $\exists$ two groups up to isomorphism of order 39. One is cyclic and other is non-abelian (see in the link for details Question on groups of order $pq$). If $G$ is cyclic then for each devisor it has unique normal subgroup. Further, if $G$ is a non-abelian group of order 39, then by using sylow theorem we can see that subgroup of the order of 13 is unique therefore it is normal. Again subgroup of order 3 can not be normal otherwise it will be unique and $G$ becomes cyclic. Therefore only one non-trivial normal subgroup in this case.

Answer 3

From your final comment let $p=3$ and $q=13$. This resolves the fact that there is a single subgroup of order $13$ which is normal, and $13$ subgroups of order $3$.

Since the possible orders of subgroups are $1, 3, 13, 39$ this leaves only the question of whether there are any normal subgroups of order $3$.

Think about one such - it is cyclic and consists of three elements $1, a, a^2$. Let $b$ be another element of order $3$. $a$ and $b$ together generate a group of order (a) greater than $3$ because it contains the distinct elements $1, a, a^2, b$; and (b) divisible by $3$ because it contains the element $a$ of order $3$. The only possibility is that $a$ and $b$ generate the whole group.

Now if the subgroup generated by $a$ were normal we'd have either:

$b^2ab=a$ (remembering that $b^3=1$) which would make $a$ and $b$ commute and generate a subgroup of order $9$ and this is impossible in a group of order $39$; or

$b^2ab=a^2$, which we can write $aba=b$ whence $abab=b^2\neq 1$. Then $(ab)^3=b^2ab=a^2\neq 1$ and $(ab)^6=1$. And $ab$ would have order $6$ (we eliminated $2$ and $3$ on the way), and this is impossible in a group of order $39$.

This is not the slickest proof, but shows it can be done with elementary calculations and without Sylow. The general proofs are better because they show more insight into the structure - and the non-abelian groups of order $pq$ have a structure it is worth understanding.

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You could also work with $c$, an element of order $13$. We have $a^2ca=c^r: r\neq 1$ because the subgroup of order $13$ is normal and the whole group is not cyclic. And that also leads (with care) to a proof that the group generated by $a$ does not have $c$ in its normaliser, and hence can't be normal.

[Note that $r$ is more restricted than shown above]