Maths › Further Pure 2 › Subgroups, Lagrange's theorem and isomorphism
Subgroups, Lagrange's theorem and isomorphism
Groups hide inside groups, and their sizes are not free: every subgroup's order divides the whole. Two groups with the same multiplication pattern are the same group in different clothing.
Builds on Groups and their axioms and Proof by induction.
IN THIS TOPIC
- Test whether a subset is a subgroup, and list the subgroups of a small group.
- Apply Lagrange's theorem to rule out impossible subgroup orders.
- Decide whether two groups of the same order are isomorphic.
WHAT YOU PROBABLY THINK
A group of order 12 could have a subgroup of order 5, since 5 is smaller than 12.
Groups inside groups
A subgroup is a subset that is itself a group under the same operation. Since associativity is inherited, only three things need checking: the identity is present, the subset is closed, and every element's inverse is inside. Every group has at least two subgroups, the identity alone and the whole group.
WORKED EXAMPLE
All the subgroups of a small group
Find every subgroup of the integers modulo 6 under addition.
The trivial ones: {0} of order 1, and the whole group of order 6.
{0, 3} is closed, since 3 + 3 = 0, and contains inverses: order 2.
{0, 2, 4} is closed with 2 + 4 = 0: order 3.
Four subgroups, of orders 1, 2, 3 and 6. Every one of those divides 6, and no subgroup of order 4 or 5 exists.
Lagrange's theorem makes that pattern a law: in a finite group, the order of any subgroup divides the order of the group. The opening claim breaks it outright, since 5 does not divide 12. The theorem also explains why every element's order divides the group's: the powers of an element form a subgroup.
Same group, different clothes
Two groups are isomorphic when their elements can be paired up so that the operation is preserved: a relabelling that turns one Cayley table into the other. Any two cyclic groups of the same order are isomorphic, so at each order the real question is how many genuinely different structures exist.
WORKED EXAMPLE
Telling two order-four groups apart
Show that {1, −1, i, −i} under multiplication and the Klein group, in which every non-identity element squares to the identity, are not isomorphic.
In the first, i has order 4: i, −1, −i, 1 takes four steps to return.
In the Klein group every non-identity element has order 2 by definition, so no element has order 4.
Isomorphism preserves the order of elements, so no relabelling can match them. Both have four elements, but they are different groups: exactly the two that exist at order 4.
YOUR TURN
Ruling things out with Lagrange
A group has order 10. List the possible orders of its subgroups, and of its elements.
Show the working
Subgroup orders must divide 10: only 1, 2, 5 and 10 are possible.
The powers of any element form a subgroup, so element orders come from the same list.
In particular no element can have order 3, 4, 6, 7, 8 or 9, however large the group looks. Lagrange rules them out without any calculation.
THE EXAM BIT
- For a subgroup test, check identity, closure and inverses; associativity is inherited and need not be proved.
- Quote Lagrange by name when using it, and state which order divides which.
- To disprove isomorphism, find a structural difference: element orders and whether the group is abelian both work.
- Isomorphism questions are restricted to groups of order at most eight, so an exhaustive check is realistic.
CHECK YOURSELF
A group has order 15. Explain why it cannot contain a subgroup of order 6.
Show a hint
Lagrange's theorem constrains subgroup orders.
Show the answer
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A subgroup needs the identity, closure and inverses; associativity comes free from the parent group.
Lagrange: every subgroup's order divides the group's order, and so does every element's order.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Test whether a subset is a subgroup, and list the subgroups of a small group.
- Apply Lagrange's theorem to rule out impossible subgroup orders.
- Decide whether two groups of the same order are isomorphic.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.