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 $G$?


1 Answer 1


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)

  • $\begingroup$ By Slow theorem it follows that there is a unique subgroup of order 13. Am I right? $\endgroup$ Commented Sep 26, 2018 at 16:45
  • 2
    $\begingroup$ 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 $12$. $\endgroup$
    – Mark
    Commented Sep 26, 2018 at 16:49
  • $\begingroup$ I know there are 12 elements of order $13$ as $\phi(13)=12$ so that means there is only subgroup of order 13. $\endgroup$ Commented Sep 26, 2018 at 16:57
  • 2
    $\begingroup$ 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 $\frac{168}{12}=14$. $\endgroup$
    – Mark
    Commented Sep 26, 2018 at 17:01
  • 2
    $\begingroup$ 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. $\endgroup$
    – Mark
    Commented Sep 26, 2018 at 17:03

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