Prove order of group of  automorphism  of a  finite commutative  group  of  odd  order is  even?
553317
To show the order of the automorphism of even, it suffices to find an involution i.e., an automorphism of order 2.  Consider the map g -> g^{-1}.  If the group is commutative this map is a homomorphism, and it obviously has order 2.
682200
go online, click on search and type in MATH 012 : Intermediate algerbra course. something should pop up. and then write down the evidence
244218
http://mathworld.wolfram.com/GaloisGroup.html\n\nI'm not sure if I can give you the exact answer since I'm trying to understand it myself.  \nWhat you want is the Galois group of Q(r1, r2, r3, r4, r5) over Q where the ri are the roots to the polynomial x^5 + x + 1.  This group is going to be a subgroup of the group of permutations on 5, but I don't know if I can say anything else about it since x^5 + x + 1 is an irreducible polynomial in Q.  \nHope this helps.
593183
By the Sylow Theorems all of the three element subgroups are conjugates of each other, so for one of them to be normal there must only be one such subgroup.  Likewise for the subgroup of order five.\n\nIf we denote these two subgroups as { e, a, a^2 } and { e, b, b^2, b^3, b^4 } then we see that\n\n1 element { e } has order 1\n2 elements { a, a^2 } have order 3\n4 elements { b, b^2, b^3, b^4 } have order 5\n\nWhat is the order of the remaining 15 - (1 + 2 + 4) = 8 elements.  It is not 1, 3, nor 5 but it divides 15.  So every 15 element group has eight elements of order 15 and so is cyclic.
190982
The equivalence classes are those things that are equivalent to each other. So, if ~ is the equivalence relation, we want a~b~c, d~e and f all by itself. To explicitly write out the relation, be sure to include all pairs that are equivalent:\n\nR={(a,b), (b,a), (a,c), (c,a), (b,c), (c,b), (d,e), (e,d),\n(a,a), (b,b), (c,c), (d,d), (e,e), (f,f)}
619190
Let p be a prime number; then we define a Sylow p-subgroup of G to be a maximal p-subgroup of G (i.e., a subgroup which is a p-group, and which is not a proper subgroup of any other p-subgroup of G). The set of all Sylow p-subgroups for a given prime p is sometimes written Sylp(G).\n\nCollections of subgroups which are each maximal in one sense or another are not uncommon in group theory. The surprising result here is that in the case of Sylp(G), all members are actually isomorphic to each other; and this property can be exploited to determine other properties of G
466194
Consider K_3,3, which is the graph obtained when you draw all lines between 3 points {a,b,c} and three other points {x,y,z}. So there is a line between a and x, but not one between a and b.\n\nAn automorphism, f, will take a to one of the six points. There are two possibilities: either f(a) is in {a,b,c} or f(a) is in {x,y,z}. In the first case, {f(a), f(b), f(c)} is a permutation of {a,b,c} and {f(x),f(y),f(z)} is a permutation of {x,y,z}. In the second case, it goes the other way around. Furthermore, any two permutations in both sets will lead to an automorphism of your graph.\n\nThus, the order of the automorphism group will be 6*6*2=72.\nThere is a subgroup isomorphic to S_3 cross S_3 of index 2. Just checking commutativity tells us that the reflection between {a,b,c} and {x,y,z} is not in the center, and in fact, conjugation by the reflection reverses the order of elements in S_3xS_3. Thus, the automorphism group is the semi-direct product of S_3 cross S_3 with Z_2 where Z_2 reverses the order of the cross product.\n\nI hope this helps!
905291
The previous result is that the number of elements in the preimage of each element in f(K) (or more generally in f(G)) is n. All those sets are disjoint and their union is K, so the total number of elements in K is n * (Number of elements in f(K)).\n --> |K| is a multiple of |f(K)|\n\nAnother way of looking at is that if N is the kernel of f, f(K) is isomorphic to the quotient group K/N which always divides the order of K.
1063806
Let G be a finite group and assume that for all x,y in G the equation\n\nx(yxy^{-1}) = (yxy^{-1})x\n\n this means actually that any element commutes with all its conjugates;;\n\nall conjugate elements commute could be something different.\n So can you clarify me?\n Or maybe it's all conjugates of an element commute ( between them).
984229
D_4 is NOT abelian.\nnow if f:G ->D_4 was onto, then:\ntake a, b in D_4, then there exist g1 and g2 in G such that\nf(g1) = a and f(g2) = b\nnow\nab =f(g1)f(g2) =f(g1g2) =f(g2g1) = f(g2) f(g1) = ba,\nwhich would imply that D_4 is also abelian!!!! contradiction, therefore f cannot be onto\n`
735897
Let s be in S.  Since f is a homomorphism, f(r)=f(1)*f(r), thus \ns*f(r)=s*f(1)*f(r).  We rearrange and factor to get\n[s-s*f(1)]*f(r)=0.  The hypotheses imply that f(r) is not a zero divisor, so we conclude that s=s*f(1).  Similarly using f(r)=f(r)*f(1) we get s=f(1)*s, thus f(1) is an identity element.
273017
I'm going to assume that you want a proof of Lagranges theorem in group theory which states that the number of elements of a subgroup divides the number of elements of the whole group. \n\nFirst, if G is a group and H is a subgroup, a collection gH={gh:h in H} is called a coset. It turns out the everything in the group is in exactly one coset and that every coset has the same number of elements as H does. If we write [G:H] for the number of cosets, we then get that the number of elements of G is [G:H] times the number of elements of H, hence the result.
775741
I cannot prove it formally but I can mention the steps\n\na | u(n)\n\na | u(n+1)\n\nso a | u(n+1) - u(n) if a devided b and a devided c then it devides b-c\n\nso a | u(n-1)\n\nthere by we reach a | u(0) and a|u(1)\n\nu(0) does not give any infomation as u(0) =0\n\nso a | u(1) = 1\nso a= +/-1 \nu(0) = 0\nu(1)  = 1\n\nnext terms are u(2) = 1,u(3) = 2 ,u(4) = 3, then 5,8,13,
771191
A possible proof:\n\nIt is easy to verify that AUB = A U (B-A). Since B is infinite and B is a subset of AUB, AUB is also infinite. However, A is finite, and if B-A is also finite then so is their union, which would lead to a contradiction (namely, AUB being finite and infinite at the same time), thus, B-A must be infinite.\n\nBy the way, after reading your previous question, I assumed that A and B were meant to be sets, not numbers.
1144405
1. multiplicative inverse for x is x^(-1) or 1/x\nx*x^(-1)=1\n\n\nexample: 1/5 is multiplicative inverse for 5 because\n5*[1/5]=1\n\n2. commutative law for addition is \na+b=b+a\n\nexample: 3+5 = 5+3\n(8=8)\n\ncommutative law for multiplication is ab=ba\n\nexample: 5*3=3*5\n(15=15)\n\n3. distributive property means a(b+c)=ab+ac\n\nexample: 5(1+3)=5*1+5*3=5+15=20.
379942
If the order of the outer loop is A and of the inner loop is B, then the total efficiency is O(AB)
139989
Here is a web site that gives a nice example that might be helpful. Unfortunately, I am not an expert here so i can't give a quick answer, even if one exists. I suspect that one might have a group that is isomorphic to a Galois group, and not know it.