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## Notes for the week of November 2, 2015

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by: Lauren Caldwell

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# Notes for the week of November 2, 2015 Math 550

Marketplace > Marshall University > Mathematics (M) > Math 550 > Notes for the week of November 2 2015
Lauren Caldwell
Marshall
GPA 3.52
Modern Algebra I
Dr. Niese

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These are my notes for Dr. Niese's Modern Algebra I course for the past week. I've typed them up, gone over them for errors, and added a couple things to clarify points that weren't immediately cle...
COURSE
Modern Algebra I
PROF.
Dr. Niese
TYPE
Class Notes
PAGES
4
WORDS
KARMA
25 ?

## 1

1 review
"I'm a really bad notetaker and the opportunity to connect with a student who can provide this help is amazing. Thank you so much StudySoup, I will be back!!!"
Dr. Kayli Nienow

## Popular in Mathematics (M)

This 4 page Class Notes was uploaded by Lauren Caldwell on Sunday November 8, 2015. The Class Notes belongs to Math 550 at Marshall University taught by Dr. Niese in Fall 2015. Since its upload, it has received 11 views. For similar materials see Modern Algebra I in Mathematics (M) at Marshall University.

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## Reviews for Notes for the week of November 2, 2015

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Date Created: 11/08/15
Modern Algebra Notes Week of 1122015 Normal Subgroups Definition Let G be a group A subgroup H of G is called normal if gH Hg for all elements g within G The symbol used for normal subgroups is similar to the subgroup symbol with the triangle of the S completed Because this writing program does not have that character I am unable to use it in these notes Example In S3 let H 1 1 2 3 1 3 2 gH H and 1 2H 1 2 2 3 1 3 Hg H and H1 2 1 2 1 3 2 3 Thus H is a normal subgroup of S3 Theorem Let I G gt G39 be a homomorphism and let a be an element of G Then aKerI KerIa Further I391Ia aKerI PLof Let I and a be as above Let X be a member of D391Ia Then IX Ia So Ia391IX e39 Since I is a homomorphism Ia391IX Ia391IX Ia391X e39 Thus a391X is a member of KerI So a391X h and thus X ah for some h in KerI So X is a member of aKerI Thus D391Ia is a subset of aKerI Now let y be a member of aKerI Then y ah for some h in KerD Then Iy Iah IaIh Iae39 Ia Thus Iy is a member of I391Ia and aKerI is a subset of I391Ia Thus D391Ia aKerI A similar proof shows D391Ia KerDa Use IXIa391 to show D391Ia is a subset of KerDa and make y a member of KerIa to show KerIa is a subset of D391Dal In essence the above theorem says that kernels of homomorphisms are normal subgroups So you can prove a subgroup is normal by finding a homomorphism whose kernel is that subgroup For example A11 is a normal subgroup of Sn since Kersgn An Theorem Let I G gt G39 be a homomorphism I is onetoone inj ective if and only if KerI e PLof Let I G gt G39 First let I be onetoone Then since Ie e39 KerI e Now let KerI e Let a and b be members of G Assume Ia Ib Then b is a member of D391Ia So b is a member of aKerI and since KerI e b a Thus I is onetoone Note We can now show that I G gt G39 is an isomorphism by showing 1 that I is a homomorphism 2 KerI e and 3 I is onto surjective In other words if we show that I is homomorphic first we can replace the onetoone step with showing KerI e Theorem If I G gt G39 is a homomorphism and G is finite then IG divides G Proof Let I G gt G39 be a homomorphism For each element a within G I391Ia aKerI Thus ltIG G KerI and ltIG 39G39AKe qm So IGKerI G and thus IG divides G Theorem Let H be a subgroup of G Leftcoset multiplication is welldefined by aH bH abG if and only if H is a normal subgroup of G PLof Suppose H is a normal subgroup of G Consider aH and bH for some elements a and b within G Then aH ath and bH bh2H where h and h2 are elements of H So aHbH abH and ah1Hbh2H ahlbh2H Note that h1b is a member of Hb Since H is a normal subgroup of G Hb bH so h1b is a member of bH Thus h1b bh3 for some h3 in H So ahlbh2H abh3h2H abH since h3h2 is a member of H Thus leftcoset multiplication is welldefined on H Now assume leftcoset multiplication is welldefined on cosets of a subgroup H of G Let a be an element of G Let g be a member of aH Then g ah for some element h in H Further since leftcoset multiplication is welldefined H aa391H aHa39lH nga39lH gal5H which implies that ga391 is a member of H Thus ga391 h for some h in H Finally g ha so g is a member of Ha Thus aH is a subgroup of Ha A similar argument shows that Ha is a subgroup of aH Thus aH Ha Theorem Let H be a normal subgroup of G The left cosets of H in G for a group called the factor group of G by H denoted GH read G mod H under the binary operation aHbH abH PLof Associativity Let aH bH and cH be elements of GH Then aHbHlCH abHCH abcH abcH aHbCH aHbHCHl Identity Consider H within GH and let aH be a member of GH Then HaH eaH aH aeH aHH Thus H is the identity of GH Inverses Let aH be a member of GH Since a is a member of G a391 is a member of G so a39lH is a member of GH Further aHa39lH aa391H H a39laH a39lHaH Thus a39lH is the inverse of aH in GH Example Z2Z 2Z 12Z Z3Z 3Z 13Z 23Z ZnZ nZ 1nZ n1nZ Note ZnZ is isomorphic to ZIl under IknZ kmod n Theorem Let I G gt G39 be a homomorphism with KerI H The map u GH gt IG defined by uaH Ia is both welldefined and an isomorphism PLof Let G G39 I and H be as above Let aH be a member of GH Then aH ahH for some element h within H Then uaH Ia IaIh since h is a member of KerI Iah since I is homomorphic uahH Therefore u is welldefined Now let aH and bH be members of GH Then uaHbH uabH Iab ltIgtalt1gtb uaHubH Thus u is homomorphic Now note that Keru H This implies u is onetoone Finally let g be a member of IG Then g Ia for some element a of G Now uaH Ia g Thus u is onto Thus u is an isomorphism Theorem Let H be a normal subgroup of G Then v G gt GH defined by yx XH for all elements X within G is a homomorphism and Kery H Fundamental Homomorphism Theorem Let I G gt G39 be a homomorphism with KerI H Then IG is a group u GH gt IG defined by ugH Ig is an isomorphism and I my where v G gt GH is defined by yg gH Example Characterize Z3XZ50XZ5 We need to find a group isomorphic to Z3XZ5OXZ5 We want a homomorphism I mapping from Z3XZ5 to this group such that KerI 0gtltZ5 because by the Fundamental Homomorphism Theorem if 1 exists Z3XZ5KCI D will be isomorphic to IZ3gtltZ5 Recall the projection map m Z3XZ5 gt Z3 defined by mxy X Compute K61 7I1 K61 7I1 0y y is a member of Z5 0gtltZ5 FUI39thCI39 7T1 Z3XZ5 Z3 By the Fundamental Homomorphism Theorem Z3XZ5OXZ5 is isomorphic to Z3 Theorem Let H be a normal subgroup of G The following are equivalent 1 H is a normal subgroup of G the symbol for normal subgroups was actually used here 2 gH Hg for all elements g in G 3 gHg391 H for all elements g in G 4 ghg391 is a member of H for all elements g in G and h in H The theorem above was not proven in class but an outline for the idea of the proof was given Proof Outline We know we have 1 if and only if we have 2 Show that 2 implies 3 implies 4 implies 2 Automorphisms Definition An isomorphism from G to G is called an automorphism Definition Fix an element g Within G Then ig G gt G defined by igx gxg391 for all elements X in G is the inner automorphism of G by g gxg391 is called conjugation of X by g Theorem Let H be a subgroup of G Then igH H for all elements g in G if and only if H is a normal subgroup of G We then say H is invariant under ig for all elements g in G Theorem Let G be a cyclic group and N a normal subgroup of G Then GN is cyclic PLof Let G and N be as above Since G is cyclic G ltagt for some element a Within G Let gN be a member of GN Since G ltagt g a11 for some integer n Thus gN anN aNan391N aNaNa 392N etc Following this logic GN ltaNgt under coset multiplication Thus GN is cyclic

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