math

You don’t have to do Question 6. Just forgot it. The question 3 have to use the format which is used in the class. I’m already posted the class note in the file. Also the homework are required to be typed in LaTeX and come out with PDF format.

MATH 421/521 Section B Intro to Abstract Algebra HW3 — Fall 2020

All homework are required to be typed in LaTeX. You can use the free online editor
http://www.overleaf.com. See https://www.overleaf.com/learn/ for a brief introduction.

HW3 is due Tuesday September 29, by 11:59pm. Please upload your solutions on canvas under
Assignments by the due time.

1. List the left cosets of the subgroups in each of the following. Here 〈a〉 denotes the subgroup
generated by the element a.

(a) 〈8〉 in (Z24,+)

(b) 〈3〉 in U(8)

(c) 3Z in Z (where 3Z = {3k : k ∈ Z}.)

(d) An in Sn (where An is the set of all even permutations on {1, . . . , n}.)

2. Find all the left cosets of H = {1, 19} in U(30).

3. Given a finite group G and H a subgroup. Using the same argument as in the Lemma proved
in class, one can show all the statements of the lemma hold analogously for right cosets of H in G.
In particular, Ha = H if and only if a ∈ H and the distinct right cosets of H form a partition of
G. Now, suppose that |H| = |G|/2.

(a) Show that for every a ∈ G, aH = Ha. (comment: in general aH and Ha are not necessarily
equal. But with our condition |H| = |G|/2 here, this indeed holds.)

(b) Suppose a, b ∈ G are two elements of G that are not in H. Prove that ab ∈ H.

4. Let G be a group of order 63. Prove that G must have an element of order 3.

5. Let G be a group of order 155. Suppose a, b are two nonidentity elements of G that have different
orders. Prove that the only subgroup of G that contains both a and b must be G itself. (hint: By
Larange’s theorem, the order of any nonidentity element must be one of 5, 31, 155. If one of a, b
has order 155 then the statement is quite easy to prove. So one may assume |a| = 5 and |b| = 31.
Consider how Theorem 7.2 might be relevant.)

6. (Graduates only) Prove that every subgroup of Dn that has an odd order must be cyclic.

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MTH4211521 Lecture 14 9121 20

Det Let G be a group H a subgroup
For eachelement a EG the left coset oft
associated with a is the set

at haha h EH’s

In additive notation
atte hath htt 9

we canthinkof a H as some sort of shift
of H by a
The right coset oft associated with a is
defined as

Ha L ha HEH 9
Reward for each a aft may not equal Ha

EI Let G Zg H 244
Then the left Cosets are

0 184 10,2 4,69 I124 21,3 5,79
2 184 42.46,09 3 124 23 5,7 IE
4 1 44 24 6 0,29 51 24 45,8 1,33
6 t Za 26,0 2.49 7124 47 I 359

Note that 0124 2124 4 124 6424 70,246
It 24 3 24 5424 7 24 21,35,79

EI Let G Sss H La B I
write down all the left cosees of H

Recall the elements of G ave

C 331 43 l 33 GD

L23 uz 1223 clad

f I 132 EI CB

LD H H
B It L 233cL 31113 1235 42334
C12 H L 112 42311339 L R 113238
123 H 411237 11233433 11237 23
432 SHE 46132 13234339 2932 112J’s
B H L 1137 4374339 Lab 4379 14

Note CDH 4 B H uh 11376
IN 123H 4 237423
2 It C1323ft iz 113239

coincidence

Lemme Properties of Cosets
Let G be a group and H a subgroup
Let a b EG Then

I a C AH
2 aH H if and only if a EH
3 CabSH albH
4 AH bH if andonly if aEbH
5 Either aH bH or at Nbt _of
6 aH bH it andonly if a b EH
7 late4171
8 att Ha if andonly if H aHaY
9 att is a subgroup of G if andonlyif

a Etl

tf 1,43 Exercises
4 First suppose a Ebt Then
a bhp for some h EH
We show attebH and bH Eat
Let X be any element of a H

Then A ah for some h E H
Now f ah bheh bChih
since it is closed hehEH so debt

Since a bhi b ahit Let y be anyelement
of BH Then Y bh for some HEM Now

y bh ahi h aChi h Gatt
where we used the fact that hith EH
Since XYarearbitrary we have ate BHand
BH Eat So at bt

5 Suppose att n BHI d Lets be an
element in aHnbH then

I ah bhz for some hihaGH
a bhahi Ebt

By 4 att BH

6 By 4 aH bH if and onl if aEbH
Now a Ebt Eh EH a bh

7 HEH b a hs b la EH
at BEM

7,8 Exercises

9 By 5 the distinct coset of H partition G
Amongthem only It contains e and is a subgroup
By 4 aH eH H if andonly if aEH

By Lemma itemsCD the distinct coset of H
form a partition of G

100002
Ee Find the distinct co sets of H 41,3 5,9 159
in G U Zz L1,3 37,9 B 15,17 19,219

Weknow
1H 3 test 9H 45 H L1,3 5,9159

Now pick an element not in H find its cosee
say 7
Ftl L 7 21,35 63 1059 mod22

47,21 13 19 179
BY the lemma

7 A 1311 1714 1914 21 HE471317,1948

So these two are the distinct coset

Thur71 Lagrange’s thus
Let G be a finite group and H a subgroupofG
Then IHI divides G Morever the number of
distinct Clefts assets of ft in Cr in 1611411
PI
By Lemma eD ta EGe a Gat
So alfgate G By E Va.BE G
either at bat or aH n bit of
So if a H Azt AmH are all the distinct
Cosets then they must be pairwise
disjointand their union is G
So they form a partition of G
By CF AiH1492141 1amHEL HM
SO 1Gt 2192171 dueto the partition

f I

m CHI

SO m IGI th B implies Hel11611h11
and that distinct assets of Hin G e E

Reward the number of left Cosets of H
in G is also called the index of Hin Gand
is denotedby 1 Git

Y 1

Cord If G B a finite groupand H a subgroup
then 1betel 161 1141

Core If G is a finite group the order lad
of any element a must divide Gl
PI sa is a cyclic subgroupof G oforder
lat By thin F l lal must dicide G

EI U122 L I 3,5 7,9 13 15,17 19,219
WED1 10 the only possible

ordersof an element are 1 2,5 or lo
For instance we saw earlier that 131 5

Cor3 Let p be a prime Let G be a group
of order p Then G must be cyclic
PI Let a be any nonidentity element
Then lat By Cor 3 lal mustdivide p
But the only divisors otp are 1 and P so
at p So G La

Cort Let G be a group of finite order n
Let a EG Then a ee
PI Let metal By Cor 3 mln So
ne m K for some integer K Now

an Cam keek e

Corte Fermat’s little Theorems
For every integer a andeveryprime p
APmod p a modp

PI By the division algorithm a mptr
where m r EE and o Er Ep 1 So
amodp r By properties of modulo
operation V xiyifxmodp y modp
henxnmudp ynmodp.fr all n EET

So it suffices to prove that
pPmed p r

consider the group VCR L1,2
n p 19

where the group operation is multiplication
modulo p By Cor 4

getmod p I

By properties of modulo operation
V a.bi.DE ifamodp bmodpand cmodp dmodp
then acmadp abdmodp
So rMmodp I made

rmodp rmodp
rPmodp rmodp h EB

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\begin{center}
{\large \bf \sc MATH 421/521 Section B Intro to Abstract Algebra HW3 — Fall 2020} \\
\vspace{0.5em}
\end{center}
All homework are required to be typed in LaTeX. You can use the free online editor\\
\url{http://www.overleaf.com}. See \url{https://www.overleaf.com/learn/}
for a brief introduction. \\
HW3 is due Tuesday September 29, by 11:59pm. Please upload your solutions on canvas under Assignments by the due time.

\medskip

1. List the left cosets of the subgroups in each of the following.
Here $\langle a\rangle$ denotes the subgroup generated by the element $a$.
\medskip
(a) $\langle 8 \rangle$ in $(\mathbb{Z}_{24},+)$
\medskip
(b) $\langle 3 \rangle$ in $U(8)$

\medskip
(c) $3\mathbb{Z}$ in $\mathbb{Z}$\quad (where $3\mathbb{Z}=\{3k: k\in \mathbb{Z}\}$.)

\medskip
(d) $A_n$ in $S_n$ \quad (where $A_n$ is the set of all even permutations on $\{1,\dots, n\}$.)
\bigskip

2. Find all the left cosets of $H=\{1,19\}$ in $U(30)$.
\bigskip
3. Given a finite group $G$ and $H$ a subgroup.
Using the same argument as in the Lemma proved in class, one can show all the statements
of the lemma hold analogously for right cosets of $H$ in $G$.
In particular, $Ha=H$ if and only if $a\in H$ and the distinct right cosets of $H$ form a partition of $G$.
Now, suppose that $|H|=|G|/2$.
\medskip
(a) Show that for every $a\in G$, $aH=Ha$. (comment: in general $aH$ and $Ha$ are not necessarily equal.
But with our condition $|H|=|G|/2$ here, this indeed holds.)
\medskip
(b) Suppose $a,b\in G$ are two elements of $G$ that are not in $H$. Prove that $ab\in H$.

\bigskip

4. Let $G$ be a group of order $63$. Prove that $G$ must have an element of order $3$.
\medskip
5. Let $G$ be a group of order $155$. Suppose $a,b$ are two nonidentity elements of $G$
that have different orders. Prove that the only subgroup of $G$ that contains both $a$ and $b$
must be $G$ itself. (hint: By Larange’s theorem, the order of any nonidentity element must be one of $5,31, 155$.
If one of $a,b$ has order $155$ then the statement is quite easy to prove. So one may assume $|a|=5$
and $|b|=31$. Consider how Theorem 7.2 might be relevant.)
\bigskip
6. (Graduates only) Prove that every subgroup of $D_n$ that has an odd order must be cyclic.

\end{document}