Possible structures for a group G where every element has order dividing 63 and in which there are 108 elements of order exactly 63.

Question

I am trying prove the problem in the title. Let $G$ be a finite abelian group in which every element has order dividing 63 and in which there are 108 elements or order exactly 63. I want to determine all the possibilities for the structure of G.

My attempt is the following: I know that since every element has order dividing 63, then the order of any element will be either 3, 9 or 7. Furthermore, by the fundamental theorem for the finitely generated abelian groups, I know that the group will be a direct product/sum of a certain number of copies of $\mathbb{Z}/3\mathbb{Z}$, $\mathbb{Z}/9\mathbb{Z}$ and $\mathbb{Z}/7\mathbb{Z}$. So my group G will be isomorphic to $(\mathbb{Z}/3\mathbb{Z})^a \oplus (\mathbb{Z}/9\mathbb{Z})^b \oplus (\mathbb{Z}/7\mathbb{Z})^c$ where $a,b,c$ are integers such that $b, c \geq 1$ (since every element has order dividing 63). Now, this is where I am stuck. Since there are 108 elements of order exactly 63, I think I want to count how many elements have order 63 in my direct sum (find an expression for the number of these elements in terms of $a,b,c$) and from there find the adequate values of $a,b,c$. Is this is a good way to approach this problem? How could I count this number of elements of order 63? Thanks!

Answer 1

Note that the element of $G = (\mathbb{Z}/3\mathbb{Z})^k \oplus (\mathbb{Z}/9\mathbb{Z})^s \oplus (\mathbb{Z}/7\mathbb{Z})^r$ are of the form $(a_1,\cdots,a_k,b_1,\cdots,b_s,c_1,\cdots,c_r)$, where $a_i,b_i,c_i$ are elements of copy of $\mathbb{Z}/3\mathbb{Z},\mathbb{Z}/9\mathbb{Z},\mathbb{Z}/7\mathbb{Z}$ respectively. Also we have that:

$$o((a_1,\cdots,a_k,b_1,\cdots,b_s,c_1,\cdots,c_r)) = \operatorname{lcm}(o(a_1),\cdots ,o(c_r))$$

From here it's easy to notice that the order would be divisible by $7$ if at least one $c_i$ has order $7$ in its respective copy of $\mathbb{Z}/7\mathbb{Z}$. Thus we have $7^r - 1$ choiches for the last $r$ coordinates s.t. the element will have an order divisible by $7$.

Similar reasoning yields that we have $9^s - 3^s$ choiches for the middle $s$ coordinates. This is true, as $\mathbb{Z}/9\mathbb{Z}$ has exactly $3$ elements of order not $9$.

Finally the choice of the first $k$ coordinates doesn't matter too much, as they alone can't contribute to an element having order divisible by $9$. Thus the number of elements of order $63$ in $G$ is exactly:

$$3^k(9^s-3^s)(7^r-1)$$

Now it's not hard to conclude that $r=1$, as if $r\ge 2$ we obtain a contradiction by:

$$108 = 3^k(9^s-3^s)(7^r-1) \ge 3^k(9-3)(7^2-1) = 288\cdot3^k $$

Similarly $s=1$, as otherwise:

$$108 = 108 = 3^k(9^s-3^s)(7-1) \ge 3^k(9^2-3^2)(7-1) = 432\cdot3^k$$

Finally we are left with:

$$108 = 3^k\cdot 6 \cdot 6 \implies k=1$$

Thus we must have $G= \mathbb{Z}/3\mathbb{Z} \oplus \mathbb{Z}/9\mathbb{Z} \oplus\mathbb{Z}/7\mathbb{Z} $

Answer 2

In $\mathbb{Z/7Z}$ there are $1$ element of order $1$ and $6$ elements of order $6$. In $\mathbb{Z/9Z}$ there are $1$ element of order $1$, $2$ elements of order $2$, and $6$ elements of order $9$. A $9$ and a $7$ give you $36$ elements of order $63$, so you need one $3$ to get $108$.

Answer 3

Here's another approach: in any finite abelian group $G$, if $g\in G$ has maximal order, then we can write $$ G = \langle g\rangle\times K $$ for some subgroup $K$.

Applying that to your case, we see we can write $$ G = C_{63}\times K $$

If there were elements of order $63$ in $K$, we could go again and write $$ G = C_{63}\times C_{63}\times H $$

but then $G$ would have more than $108$ elements of order $63$. Thus $ G = C_{63}\times K$ and $G$ then has $36*|K|$ elements of order $63$. Thus $K$ has order $3$ and you're done.