Rayat Shikshan Sanstha’s S.M.Joshi College, Hadapsar -28

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RAYAT SHIKSHAN SANSTHA’S S.M.JOSHI COLLEGE HADAPSAR, PUNE
RAYAT SHIKSHAN SANSTHA’S S. M. JOSHI COLLEGE HADAPSAR, PUNE
RAYAT SHIKSHAN SANSTHA’S S.M.JOSHI COLLEGE, HADAPSAR, PUNE
RAYAT SHIKSHAN SANSTHA’S S. M. JOSHI COLLEGE HADAPSAR, PUNE
Rayat Shikshan Sanstha’s S.M.Joshi College, Hadapsar -28
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RAYAT SHIKSHAN SANSTHA’S S.M.JOSHI COLLEGE HADAPSAR, PUNE
Rayat Shikshan Sanstha’s S.M.Joshi College, Hadapsar -28
Rayat Shikshan Sanstha’s S.M.Joshi College, Hadapsar -28
Rayat Shikshan Sanstha’s S.M.Joshi College, Hadapsar -28
Rayat Shikshan Sanstha’s S.M.Joshi College, Hadapsar -28
Rayat Shikshan Sanstha’s S.M.Joshi College, Hadapsar -28
Rayat Shikshan Sanstha’s S.M.Joshi College, Hadapsar -28
Rayat Shikshan Sanstha’s S.M.Joshi College, Hadapsar -28
Rayat Shikshan Sanstha’s S.M.Joshi College, Hadapsar -28
Rayat Shikshan Sanstha’s S.M.Joshi College, Hadapsar -28
RAYAT SHIKSHAN SANSTHA’S S.M.JOSHI COLLEGE HADAPSAR, PUNE
Rayat Shikshan Sanstha’s S.M.Joshi College, Hadapsar -28
RAYAT SHIKSHAN SANSTHA’S S.M.JOSHI COLLEGE HADAPSAR, PUNE
Rayat Shikshan Sanstha’s S.M.Joshi College, Hadapsar -28
RAYAT SHIKSHAN SANSTHA’S S.M.JOSHI COLLEGE HADAPSAR, PUNE
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Rayat Shikshan Sanstha’s S.M.Joshi College, Hadapsar -28 Department of Mathematics Power Point Presentation Topic – Finite Groups & Subgruops Prof. Darekar S.R

Finite Groups & Subgroups

Order of a group Definition: The number of elements of a group (finite or infinite) is called its order. Notation: We will use |G| to denote the order of group G.

Examples |D4| = |Dn| = |<R90>| = |Zn| = |U(8)| = |U(11)| = |Z| = 8 2n 4 n 10 ∞

Order of an element Definition: The order of an element g in a group G is the smallest positive integer n such that gn = e (In additive notation, ng = 0). If no such integer exists, we say g has infinite order. Notation: The order of g is denoted |g|.

Examples In D4, |R90| = In D4, |H| = In Z10, |4| = In Z11, |4| = In U(8), |5| = In U(9), |5| = In Z, |1| = 4 ( R490 = R0) 2 ( H2 = R0) 5 (5•4 mod 10 = 0) 11 (11•4 mod 11 = 0) 2 (52 mod 8 = 1) 6 {5, 7, 8, 4, 2, 1} ∞ (n•1 ≠ 0 for n>0)

Group G (•mod 35) |G| = e = |5| = |10| = |15| = |20| = |30| = 6 15 1 2 3 • 5 10 15 20 25 30

Subgroups Definition: If a subset H of a group G is itself a group under the operation of G, then we say that H is a subgroup of G.

Notation We write H ≤ G to mean H is a subgroup of G. If H is not equal to G, we write H < G. We say H is a proper subgroup of G. {e} is called the trivial subgroup. All other subgroups are nontrivial.

• R0 R90 R180 R270 H V D D'

• R0 R90 R180 R270 H V D D'

• R0 R90 R180 R270 H V D D'

• R0 R90 R180 R270 H V D D'

• R0 R90 R180 R270 H V D D'

• R0 R90 R180 R270 H V D D' Note: The operation is associative, and this subset has identity and inverses The subset is not closed under *.

Subgroup tests Three important tests tell us if a nonempty subset of a group G is a subgroup of G. One-Step Subgroup Test Two-Step Subgroup Test Finite Subgroup Test

One-Step Test Let H be a nonempty subset of a group G. If ab-1 belongs to H whenever a and b belong to H, then H is a subgroup of G. (In additive groups: If a–b belongs to H whenever a and b belong to H, then H ≤ G.)

To use the One-Step Test Identify the defining property P that distinguishes elements of H. Prove the identity has property P. Assume that two elements have property P Show that ab-1 has property P. Then by the one-step test, H ≤ G. H≠ a,b in H ab-1 in H

Example - 1 Prove: Let G be an Abelian group with identity e. Let H = {x |x2 = e}. Then H ≤ G. Proof: - e2 = e, so that H is nonempty. Assume a, b in H. Then (ab-1)2 = a(b-1a)b-1 = aab-1b-1 (G is Abelian) = a2b-2 = a2(b2)-1 = ee-1 (since a and b in H) = e. By the one-step test, H ≤ G.

Example -2 Prove: The set 3Z = {3n | n in Z} (i.e. the integer multiples of 3) under the usual addition is a subgroup of Z. Proof: 0 = 3•0, so 3Z is not empty. Assume 3a and 3b are in 3Z. Then 3a – 3b = 3(a–b) is in 3Z. By the One-Step test, 3Z ≤ Z.

Two Step Test Let H be a nonempty subset of group G with operation *. If (1) H is closed under * and (2) H is closed under inverses, then H ≤ G Proof: Assume a and b are in H. By (2), b-1 is in H. By (1) ab-1 is in H. By the one-step test, H ≤ G. Sometimes it is easier to verify in two steps than in one.

Finite Subgroup Test Let H be a nonempty finite subset of a group G. If H is closed under the operation of G, then H ≤ G. Proof. Choose any a in H. By the two step test, it only remains to show that a-1 is in H.

Definition Let a be an element of a group G. The cyclic group generated by a, denoted <a> is the set of all powers of a. That is, <a> = {an | n is an integer} In additive groups, <a> = {na | n is a integer}

<a> is a subgroup Let G be group, and let a be any element of G. Then <a> is a subgroup of G. Proof: a is in <a>, so <a> is not empty. Choose any x = am and y = an in <a>. xy–1= am(an)-1 = am-n which belongs to <a> since m–n is an integer. By the one-step test, <a> is a subgroup of G.

Example - 1 • 5 10 15 20 25 30 <25> =

Example • 5 10 15 20 25 30 <25> = {25, 30, 15} Check the finite subgroup test!

Thank You