$G$ is a commutative group of order 72, which is a product of cyclic groups. What is max order of element?

Question

I'm trying to understand the following practice question that has the given answer. Can someone help? Here are some specific questions:

• I presume the notation $(\mathbf{Z}/2)$ refers to some cyclic group of order $2$ that is isomorphic to the group of integers modulus 2 such that $(\mathbf{Z}/2) \cong \{0, 1\}$?
• I understand the prime factorization of 72 is $2^3 3^2$. How do we know that an arbitrary abelian group of order $72$ is composed of the cross product of three cyclic groups of order $2$ and two cyclic groups of order $3$? What rules out the possibility of an abelian cyclic group based on a single element $x$ with order $72$ so that $G = \{ 1, x, x^2, \ldots, x^{71} \}$?
• What does this equation mean? $(\bar{6a_1}, \bar{6a_2}, \bar{6a_3}, \bar{6b_1}, \bar{6b_2}) = 0$
• Why are there three elements in $\mathbf{Z}/2$ such that $\bar{a_1}, \bar{a_2}, \bar{a_3} \in \mathbf{Z}/2$. Isn't $(\mathbf{Z}/2) \cong \{ 0, 1 \}$ with two elements?

QUESTION

Suppose $G$ is a commutative group of order 72, which is a product of cyclic groups (later we will show that all finite commutative groups are isomorphic to products of cyclic groups). What is the maximum order an element of $G$ is guaranteed to have? Give an example of a group for which this is the maximum possible order. Deduce that this group is not cyclic.

ANSWER

The maximum order guaranteed is 6: indeed, since $G$ is a product of cyclic groups, one factor must have order a multiple of 3 and one factor must have order a multiple of 2, so we must have elements of order 3 and 2; their product (since the elements commute) has order 6 (this was on a homework, but we can see explicitly: if $g$ has order three and h has order two, then $(gh)^6 = g^6h^6 = 1$ since $g$ and $h$ commute, but $(gh)^3 = g^3h^3 = h$ and $(gh)^2 = g^2h^2 = g^2$ are not the identity; thus $gh$ has order six). On the other hand, the group $(\mathbf{Z}/2)^3 \times (\mathbf{Z}/3)^2$ has order 72 but all elements have order 1, 2, 3, or 6 (because $6(\bar{a_1}, \bar{a_2}, \bar{a_3}, \bar{b_1}, \bar{b_2}) = (\bar{6a_1}, \bar{6a_2}, \bar{6a_3}, \bar{6b_1}, \bar{6b_2}) = 0$ for all $\bar{a_1}, \bar{a_2}, \bar{a_3} \in \mathbf{Z}/2$ and $\bar{b_1}, \bar{b_2} \in \mathbf{Z}/3)$.

I remark that the same argument (both paragraphs above) works to show that the maximum guaranteed order in a finite abelian group of any size is the product of the prime factors of the size (here, 2 and 3).

Answer 1

first of all you are right about your presumption $Z/2 = \mathbb{Z}/2\mathbb{Z}$, so indeed has only two elements. Similarly $Z/3 = \mathbb{Z}/3\mathbb{Z}$. On your second question, we do not know that, because it is not true. Your example is perfectly fine (It is exactly $\mathbb{Z}/72\mathbb{Z}$), but that is not the point here. What the question asks is "what maximal order is guaranteed for any commutative group of order 72".

To show this we first need to show that 6 is always an order. Then we use $(Z/2)^3 \times (Z/3)^2$ as an example of commutative group of order 72 with the maximal order being 6. Therefore, even though there may be some such groups with higher maximal order (such as your example), there is at least one such group with maximal order exactly 6. We conclude that the maximal order guaranteed for any commutative group of order 72, is indeed 6.

If the question was "what is the maximal order for an element in a commutative group of order 72", then the answer would just be 72, exactly given by your example.

Thirdly, $6a_1$ is just short for $a_1+a_1+a_1+a_1+a_1+a_1$, as usual. They should maybe have written $$(6a_1,\dots,6b_2) = (0,0,0,0,0)$$ which is the identity element (i.e. the $0$) of $(Z/2)^3 \times (Z/3)^2$.

Finally, they just write $(a_1,\dots,b_2)$ for a general element of the group, but the $a_i$ and $b_j$ may very well coincide, for example $(0,1,0,2,2)$, such that $a_1 = a_3$ and $b_1=b_2$. The point is that whatever elements you choose, multiplying by 6 gives you 0 (as in the equality above).

Answer 2

$\mathbb Z_2\times \mathbb Z_2\times \mathbb Z_2\times \mathbb Z_3 \times \mathbb Z_3$ has max order $6$. Every group must have an element of order $6$.

Proof: If one of the factors have order multiple of $6$ you are done (Since cyclic subgroups have elements of each order).

If not one factor $F_1$ must have order multiple of $2$ and another $F_2$ must have order multiple of $3$. Take $f_1\in F_1$ with order multiple of $2$ and $F_2$ with order multiple of $3$. Take an element of $G$ that has $f_1$ and $f_2$ in some coordinates. this element has order multiple of $6$.

Answer 3

Yes, the notation $(\mathbf{Z}/2)$ refers to the cyclic group of order 2.

The answer does not claim that an arbitrary abelian group of order 72 is a product of three cyclic groups of order 2 and two cyclic groups of order 3? It merely points out that such a group exists and uses it to show that the maximum order guaranteed can be no larger than 6. The other (first) part of the proof shows the opposite inequality, namely that an element of at least order 6 is guaranteed to exist.

In general, in a group written additively, $6x$ mean $x+x+x+x+x+x$. In a direct product of groups, the operation works component-wise, so in any direct product of five groups, we would have $(x_1,x_2,x_3,x_4,x_5)+(y_1,y_2,y_3,y_4,y_5)=(x_1+y_1,x_2+y_2,x_3+y_3,x_4+y_4,x_5+y_4)$, and you could use this fact repeatedly to show that $6(x_1,x_2,x_3,x_4,x_5)=(6x_1,6x_2,6x_3,6x_4,6x_5)$. In this case they use the notation $\overline{a_i}$ to refer to an element in $\mathbf{Z}/2$; $a_i$ refers to a representative in $\mathbf{Z}$, and the bar over it indicates the projection $\mathbf{Z}\to\mathbf{Z}/2$.

No assumption is made about $\overline{a_1},\overline{a_2},\overline{a_3}$ being distinct; indeed, that would be impossible, as you point out. Instead, what they are showing is that every elements $x \in (\mathbf{Z}/2)^3\times (\mathbf{Z}/3)^2$ satisfies $6x=0$, i.e. has order at most 6. And to do this, they write a general element of $(\mathbf{Z}/2)^3\times (\mathbf{Z}/3)^2$ as $(\overline{a_1},\overline{a_2},\overline{a_3},\overline{b_1},\overline{b_2})$ show that $6(\overline{a_1},\overline{a_2},\overline{a_3},\overline{b_1},\overline{b_2})=0$.