Hw9 Problem 6

Show that there are two Abelian groups of order $108$ that have exactly four subgroups of order $3$.

Solution

Observe that $108 = 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3$
By the Fundamental Theorem of Finite Abelian Groups, two abelian groups that have order 108 are
$\mathbb{Z}_9 \times \mathbb{Z}_3 \times \mathbb{Z}_4$ and $\mathbb{Z}_9 \times \mathbb{Z}_3 \times \mathbb{Z}_2 \times \mathbb{Z}_2$

First, consider $\mathbb{Z}_9 \times \mathbb{Z}_3 \times \mathbb{Z}_4$
We desire to find all subgroups of order $3$.
By a theorem, for $(a,b,c) \in \mathbb{Z}_9 \times \mathbb{Z}_3 \times \mathbb{Z}_4, \mid (a,b,c) \mid = lcm( \mid a \mid, \mid b \mid, \mid c \mid) = 3$
Thus, since $3$ is a multiple of $1$ and $3$, it must be the case that $\mid a \mid, \mid b \mid,$ and $\mid c \mid$ are equal to $1$ or $3$.
Now, by another theorem, for $a \in G$ with $\mid G \mid = n, \mid \langle a ^s \rangle \mid = \dfrac{n}{gcd(s,n)}$
Thus, $\mid a \mid = \dfrac{9}{gcd(9,a)} , \mid b \mid = \dfrac{3}{gcd(3,b)},$ and $\mid c \mid = \dfrac{4}{gcd(4,c)}$
Hence, $\mid a \mid = 3$ is achieved when $a = 3$ or $6$; $\mid a \mid = 1$ is achieved when $a = 0$.
Also, $\mid b \mid = 3$ is achieved when $b = 1$ or $b=2$ ; $\mid b \mid = 1$ is achieved when $b = 0$.
Additionally, $\mid c \mid \not = 3 \ \ \forall \ c \in \mathbb{Z}_4$ ; $\mid c \mid = 1$ is achieved when $c = 0$.
It follows that the only possible elements of order $3$ are $(3, 0, 0) , (6, 0, 0) , (6, 1, 0) , (3, 2, 0), (3,1,0), (6,2,0), (0,2,0)$ and $(0, 1, 0)$.
However, every subgroup of order 3 contains two elements of order three. Since there are eight elements of order three, there are four distinct subgroups of order three.
Therefore, there are exactly four subgroups of order $3$.

The same argument may be applied to show that $\mathbb{Z}_9 \times \mathbb{Z}_3 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ has exactly $4$ subgroups of order $3$.

Therefore, there are $2$ groups of order $108$ with exactly $4$ subgroups of order $3$.

You should convince yourself that the other groups of order 108 do not have exactly four subgroups of order 3.