FP2 1.3 - Orders of elements and subgroups | Edexcel A-Level Further Mathematics

Find the order of a group and of an element, and recognise subgroups (FP2 1.3). You will locate the first return to the identity, test a subset using the original operation and construct the subgroup generated by one element.

Count the group and find an element's first return

How can a $2\times2$ matrix have order $4$ while belonging to a group of order $8$ ? The word “order” is doing three different jobs. Matrix dimensions, group size and repeated action must be kept separate.

You already know

A group has an identity and an inverse for every element. A Cayley table records the stated operation. Powers mean repeated group operation: for additive groups, use repeated addition. A generator reaches every group element by its powers.

For example, $U_7=\{1,2,3,4,5,6\}$ , under multiplication modulo $7$ , has six elements. Multiplying a particular member repeatedly can return to $1$ sooner than six steps.

Group order and element order

The order of a finite group $G$ is its number of elements, written $|G|$ .

The order of an element $a$ is the least positive integer $d$ with

a^d=e

We write $\operatorname{ord}(a)=d$ . If no such positive integer exists, $a$ has infinite order. In an additive group, find the least positive $d$ for which $da=0$ under its addition.

Minimality matters: finding one power equal to the identity is not enough if an earlier positive power does the same. The identity itself has order $1$ , because $e^1=e$ ; exponent zero is not a positive candidate.

Worked example 1: a return after three factors

Question. In the multiplicative group $U_7$ , determine $|U_7|$ and the order of $2$ .

Plan. Count the allowed elements, then list powers of $2$ until the identity first appears.

\begin{aligned} |U_7|&=6&&\text{six distinct group elements}\\ 2^1&\equiv2\pmod7&&\text{not the identity}\\ 2^2&\equiv4\pmod7&&\text{not the identity}\\ 2^3&\equiv8\equiv1\pmod7&&\text{first positive return} \end{aligned}

Why this step? The first return, rather than the group's number of members, determines this element's order.

Answer. $|U_7|=6$ and $\operatorname{ord}(2)=3$ .

Check. $2^6\equiv1\pmod7$ too, but $6$ is not the least positive exponent.

Change only the element from $2$ to $3$ , keeping the group and operation fixed.

Worked example 2: an element that reaches the whole group

Question. Determine the order of $3$ in $U_7$ .

Plan. Multiply by $3$ repeatedly and check all earlier powers before claiming an order.

\begin{aligned} 3^1&\equiv3\pmod7&&\text{first power}\\ 3^2&\equiv2\pmod7&&\text{multiply and reduce}\\ 3^3&\equiv6\pmod7&&\text{continue}\\ 3^4&\equiv4\pmod7&&\text{continue}\\ 3^5&\equiv5\pmod7&&\text{continue}\\ 3^6&\equiv1\pmod7&&\text{first return to the identity} \end{aligned}

Why this step? None of the preceding five powers is $1$ , so the return at six is minimal.

Answer. $\operatorname{ord}(3)=6$ . Its powers include every member, so $3$ generates $U_7$ .

Check. The six residues $3,2,6,4,5,1$ are distinct.

An element and its inverse have the same order: reversing a return cycle gives a cycle of the same least length. The order definitions and this consequence are not in the formula booklet.

Watch out

“ $2^6\equiv1\pmod7$ , therefore $2$ has order $6$ ” misses minimality. Since $2^3\equiv1\pmod7$ , its actual order is $3$ . Always check the first positive return.

Keep the ambient group fixed and try another element.

Test a subset under the same operation

A subgroup is a smaller group living inside a known one. You keep the original operation; you do not invent a new operation to make a chosen subset work.

Subgroup

A nonempty subset $H\subseteq G$ is a subgroup if it forms a group under the operation inherited from $G$ .

Associativity is inherited. Check that products stay in $H$ , that the original identity is in $H$ , and that the inverse of every member is in $H$ .

The sets $\{e\}$ and $G$ are always subgroups. A proper subgroup differs from $G$ ; the empty set is never a subgroup.

Why must the identity be the original one? Writing the operation multiplicatively, if $h\in H$ and $e_H$ were its subgroup identity, then $e_Hh=h$ ; cancelling $h$ in the ambient group gives $e_H=e$ . The inherited identity and inverses are not optional choices.

Worked example 3: a subgroup of modular addition

Question. In $G=\{0,1,\ldots,11\}$ under addition modulo $12$ , show that $H=\{0,4,8\}$ is a subgroup. State its order.

Plan. Use a small table for closure, then check identity and inverses; associativity comes from $G$ .

\begin{aligned} 4+4&\equiv8\pmod{12}&&\text{a product stays in }H\\ 4+8&\equiv0\pmod{12}&&\text{an inverse pair}\\ 8+8&\equiv4\pmod{12}&&\text{another closure check} \end{aligned}

$+_{12}$	$0$	$4$	$8$
$0$	$0$	$4$	$8$
$4$	$4$	$8$	$0$
$8$	$8$	$0$	$4$

Why this step? The full table checks every ordered pair in this finite subset, rather than just a few selected pairs.

Answer. The table proves closure; $0\in H$ is the identity, $0$ is its own inverse and $4,8$ are inverses. Associativity is inherited. Thus $H$ is a subgroup with $|H|=3$ .

Check. Repeatedly adding $4$ gives $4,8,0$ , exactly the three elements.

Change only the subset by removing $8$ , leaving $K=\{0,4\}$ .

Worked example 4: identity membership is not enough

Question. Is $K=\{0,4\}$ a subgroup of the same group $G$ ?

Plan. Test closure with the remaining nonidentity element.

\begin{aligned} 4,4&\in K&&\text{permitted inputs}\\ 4+4&\equiv8\pmod{12}&&\text{the inherited operation}\\ 8&\notin K&&\text{closure fails} \end{aligned}

Why this step? Containing the identity cannot repair a product that leaves the proposed subgroup.

Answer. No. $K$ is not a subgroup.

Check. Its element $4$ also lacks its required inverse $8$ inside $K$ .

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FP2 1.3 - Orders of elements and subgroups

Count the group and find an element's first return

Test a subset under the same operation

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