Maths › Further Pure 2 › Subgroups, Lagrange's theorem and isomorphism
Subgroups, Lagrange's theorem and isomorphism
A group can contain smaller groups, and their sizes are constrained. Lagrange's theorem states that the order of a subgroup divides the order of the group. Two groups with the same multiplication structure are isomorphic.
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.
COMMON MISCONCEPTION
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. Associativity is inherited, so only three things need checking. Is the identity present, is the subset closed, and is every element's inverse inside it. Two subgroups come free in every group, the identity on its own and the whole group. Those two are distinct in any group of order greater than 1, so such a group has at least two subgroups. In the trivial group they are the same subgroup and there is only one.
WORKED EXAMPLE
All the subgroups of a small group
Find every subgroup of the integers modulo 6 under addition.
The trivial ones are {0} of order 1, and the whole group of order 6.
{0, 3} is closed, since 3 + 3 = 0, and it contains inverses, so it is a subgroup of order 2.
{0, 2, 4} is closed with 2 + 4 = 0, giving order 3.
Four subgroups, of orders 1, 2, 3 and 6. Every one of those divides 6, and nothing of order 4 or 5 exists.
Lagrange's theorem turns that pattern into a law. In a finite group, the order of any subgroup divides the order of the group. A group of order 12 therefore has no subgroup of order 5, since 5 does not divide 12. The theorem also explains why every element's order divides the group's, because the powers of an element form a subgroup all by themselves.
Same group, different clothes
Two groups are isomorphic when their elements can be paired up so that the operation is preserved, which amounts to 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, since i, −1, −i, 1 takes four steps to return.
In the Klein group every non-identity element has order 2 by definition, so nothing has order 4.
Isomorphism preserves the order of elements, so no relabelling can match them. Both have four elements and they are different groups, and those two are the only groups of order 4 there are.
GUIDED PRACTICE
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, so only 1, 2, 5 and 10 are possible.
The powers of any element form a subgroup, so element orders come from the same list.
No element can have order 3, 4, 6, 7, 8 or 9, however large the group looks. Lagrange rules them out with no calculation at all.
ASSESSMENT FOCUS
- For a subgroup test, check identity, closure and inverses; associativity is inherited and needs no proof.
- Quote Lagrange by name when you use it, and state which order divides which.
- To disprove an isomorphism, find a structural difference. Element orders and whether the group is abelian both work.
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
By Lagrange's theorem the order of any subgroup divides 15, and the divisors are 1, 3, 5 and 15. Since 6 does not divide 15, no such subgroup exists.
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.
Or read them with their worked answers on the subgroups, lagrange's theorem and isomorphism questions page.
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.
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