# How many normal subgroups does a group of order 169 has?

How many normal subgroups does a group of order 169 has?

My efforts

First I proved a important result that any group of order $$p^2$$ is Abelian. In our case $$p=13$$

In an Abelian group every subgroup is normal. So I just have to find all the subgroups of $$G$$.

Now I know this result which goes by the name Cauchy Theorem

Cauchy's theorem is a theorem in the mathematics of group theory, named after Augustin Louis Cauchy. It states that if G is a finite group and p is a prime number dividing the order of G (the number of elements in G), then G contains an element of order p.

So I know there is an element of order $$13$$ in $$G$$ so I have a subgroup of order $$13$$ let's call it $$H$$ and since $$G$$ is Abelian $$H$$ is normal.

Obviously $$\{e\}$$ and $$G$$ are normal subgroups.

What are other normal subgroups of $$G$$?

Hint: $$\mathbb{Z_{169}}$$ and $$\mathbb{Z_{13}}\times\mathbb{Z_{13}}$$ are all the groups of order $$169$$ up to isomorphism. How many elements of order $$13$$ each one of these groups have? (note that every group of order $$13$$ is cyclic)

• By Slow theorem it follows that there is a unique subgroup of order 13. Am I right? Commented Sep 26, 2018 at 16:45
• No. Sylow theorems will not really help you here, because $169=13^2$ so the Sylow subgroups of $G$ are subgroups of order $13^2$ which can be only $G$ itself. This is why I suggest to count how many elements or order $13$ each of the two groups of order $169$ has. $13$ is a prime number so every subgroup of order $13$ must contain $12$ elements of order $13$, and every two different subgroups of order $13$ intersect trivially. So the number of subgroups of order $13$ is the number of elements of order $13$ divided by $12$.
– Mark
Commented Sep 26, 2018 at 16:49
• I know there are 12 elements of order $13$ as $\phi(13)=12$ so that means there is only subgroup of order 13. Commented Sep 26, 2018 at 16:57
• Let's take a look. In $\mathbb{Z_{169}}$ the elements of order $13$ are the non zero multiples of $13$, which are $13,26,39,...,12\times 13$. Other non trivial elements have order $169$. So here there are really $12$ elements of order $13$, which means one subgroup of order $13$. However, in $\mathbb{Z_{13}}\times\mathbb{Z_{13}}$ every non trivial element has order $13$, which means there are $168$ elements of order $13$ in that group, and hence the number of subgroups of order $13$ is $\frac{168}{12}=14$.
– Mark
Commented Sep 26, 2018 at 17:01
• So if we conclude, $\mathbb{Z_{169}}$ has $3$ normal subgroups (together with itself and $\{e\}$) while $\mathbb{Z_{13}}\times\mathbb{Z_{13}}$ has $16$ normal subgroups.
– Mark
Commented Sep 26, 2018 at 17:03