I have a set {0, 1, 2, 3, 4, 5}. What properties do I need to show to prove that it is a group under addition modulo 6? Any examples would be greatly appreciated!
123173
The set $Z_n = \{ 0, 1, . . . , n-1 \}$ for $n ≥ 1$ is a group under addition modulo n. For any $j &gt; 0$ in $Z_n$, the inverse of $j$ is $n - j$. This group is usually referred to as the group of integers modulo n.  My understanding of this is as following, please let me know if I have gotten the main idea.  Let's pick $Z_3$ as an example.  Then the members of this group is $0,1,2$ let $j=2$ then $n-j=3-2=1$  it says the inverse of $j$ is $n - j$, by definition, an inverse in a group is such ab=identity. And here the identity is 0 under addition.   therefore, 2+1=3 and 3 mod 3 =0 which is the identity.   This applies to any n in $Z_n$  So, it satisfies as a group?
87547
In the , the key $k = (a, b)$ should have the following property, i.e. GCD of $a$ and the modulus should be 1:     $\gcd (a, m) = 1$, where $m$ is the set of possible moduli   My question is, why it is required that $a$ should have the above property? What is the significance of this? And why is $b$ excluded from this condition?
49783
So I'm asked to prove that $$S_{3}\cong D_{3}$$ where $D_3$ is the dihedral group of order $6$. I know how to exhibit the isomorphism and verify every one of the $6^{2}$ pairs, but that seems so long and tedious, I'm not sure my fingers can withstand the brute force. Is there a better way?
59119
Let us say I have to find all the generators for modulo $p=7$. That must mean that:  $$\mathbb{Z}_7 = \mathbb{Z}^*_7 = \{1,2,...,7-1\}$$  So now I need to get all the generators for $7$. Now I choose randomly from the group $\mathbb{Z}_7$ and pick the number $3$. So if $3^n$ for $n = \{1,2,\dotsc,7-1\}$ can generate all elements from $\mathbb{Z}_7$, the number is considered a generator.  $$3^1 \pmod 7\equiv 3\\ 3^2 \pmod 7\equiv 2\\ 3^3 \pmod 7\equiv 6\\ 3^4 \pmod 7\equiv 4\\ 3^5 \pmod 7\equiv 5\\ 3^6 \pmod 7\equiv 1$$  Now I have found one generator. Someone claimed one can find all generators in the group with a faster method, when one already has one generator. Can someone please show me how that works?
114150
The problem is to define all subgroups of $(\mathbb Z_n,+), n \in \mathbb N$. My guess is if n is prime number, then there is only trivial subgroups. If n is not prime, then I can factorize it, and every prime divisor will generate it's own subgroup in $(\mathbb Z_n,+)$. That means that $(\mathbb Z_n,+) \cong (\mathbb Z_h,+) \times (\mathbb Z_k,+) \times \dots$ , $h,k \in \mathbb N$ are the prime factors of $n$. The problem is that I don't know how to prove it. It's pretty easy to show that, for example, in $(\mathbb Z_6,+)$ $ [2]_6$ and $[3]_6$ generate their own subgroups and $[1]_6$ generate entire $(\mathbb Z_6,+)$, but I don't know how to show that $[5]_6$ do the same, except by showing it all: $[5]^2_6 = [4]_6$ and so on.
111798
If $G$ is a group of order $6$. How do I show that there is some $x\in G$ with $x^2\neq e$?. I was thinking of trying a contra positive approach, by proving that if there is some $x^2=e$ then $G$ is Abelian. But how would I continue after showing that $x^2=e$?
117005
Are the groups $\mathbb{C}$ and $\mathbb{R}$ isomorphic under addition?  And how could I prove this ?  What about $\mathbb{Q}$ and $\mathbb{Q}[i]$ ?
1549
I have a hard time formulating proofs.  For this problem, I  can see that if $n$ is equal to $8,$ this statement is true.  $(\mathbb{Z}/8\mathbb{Z})^{\times}$ includes elements:  $1,3,5,7$, and all of these are roots of $1-x^2 \pmod 8.$  And obviously $8$ divides $24.$  But how do I prove this without depending on number calculations and only using theorems?  Help Please?  I need a step by step walk through of how to do this proof and what theorems would be appropriate to use.
67836
I am attempting to understand and compute: $(\mathbb{Z}_4 \times \mathbb{Z}_6)/\langle(0,2)\rangle$   I know $(0,2)$ generates $H = \{(0,0),(0,2),(0,4)\}$, which has  order 3 because there are 3 elements. Now, I must find all the cosets  which there should be 8. The reason there should be 8 is because $4\cdot 6/3 = 8$.  This is where I run into the issue. I have a hard time generating the cosets, but furthermore, once I actually do get the cosets, I have a hard time analyzing them and knowing what it means. If someone could walk me through a step-by-step solution that would be great.  The answer is that the factor group is isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_2$.   Reference: Fraleigh, A First  Course in Abstract Algebra, p. 147, Example 15.10.
102662
Find and prove a formula for $1\pmod{3}\cap 1\pmod{7}$.   I have tried looking at similar problems where $A$ is a subset of $B$ iff $A \cap B =A$. But it does not seem to work.   What type of formula should I be looking for and proving?
64228
Let $G$ be a group and $a,b \in G$ with $a \ne 1$ and $b \ne 1$. Assuming $ab^2 = b^3a$ and $a^2=1$ I need to prove that the order of $b$ is $5$.   I have proved by contradiction that it can't be 2 or 3 but I don't know how to prove that it must be 5 and it can't be 4.
36455
I'm having some trouble figuring out how to give a direct proof using the definitions of mod m and congruence modulo m, without using any theorem involving mod or congruence.  if $a, b, c$ &amp; $m$ are integers such that $m \geq 2$, $c &gt; 0$ and $a \equiv b \pmod{m}$, then $ac \equiv bc\pmod{mc}$
54300
Yeah so I'm kind of stuck on this problem, and I have two questions.   Is there a way to define a number mathematically so that it cannot be a multiple of $7$? I know $7k+1,\ 7k+2,\ 7k+3,\ \cdots$, but that will take ages to prove for each case.  Is this a proof? $$(7k + a)^3 \equiv (0\cdot k + a)^3 \equiv a^3 \bmod 7$$    Thank you.
31385
I need to show that there are only 2 groups of order 6 up to isomorphism.  I did prove it, but the proof is quite cumbersome. I wonder if there is a very concise proof.  My proof outline: Suppose $G$ is of order 6 and is not $\mathbb{Z}_6$. Then any element that is not identity must have order 2 or 3.  I went on to show that there must be a element of order 2 and another one of order 3, and the intersection of the cyclic subgroups generated by the two is the identity. Then I'm able to show it is isomorphic to $S_3$.
10580
Let $G$ be a finite, nonempty set with an operation $*$ such that   $G$ is closed under $*$ and $*$ is associative Given $a,b,c \in G$ with $a*b=a*c$, then $b=c$. Given $a,b,c \in G$ with $b*a=c*a$, then $b=c$.   I want to prove that $G$ is a group, but I don't know how to show that there exists an identity $e\in  G$ such that $e*x=x$ and $x*e=x$ $\forall x \in G$. I also don't know how to show that $\forall$ x $\in G$ there exists a $y \in G$ such that $y*x=e$ and $x*y=e$. How do I do this?
52208
If $G$ has at least $7$ elements of order $7$, then prove that $G$ has at least $12$ elements of order $7$.   I know that an element having order $7$ means that for any $a$ in $G$, $a^7=e$ where $e$ is the identity element. The inverses also have same order, so I thought maybe of the $7$ elements, $6$ had their inverse as well. However, then this would just mean there had to be $8$ elements, not $12$ for the $7$th element to have it's inverse.
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$G=Z_{10}\times Z_{15}$ Then $G$ contains exactly    one element of order $2$ five element of order $3$ 24 elements of order $5$ 24 elements of order $10$   what I think is 1 is correct, not able to guess others.