Describe explicitly all homomorphisms h: C_3 --->Aut(C_13). Hence describe all groups of order 39. How many such groups are there , up to isomorphism?
What do they mean by up to isomorphism?.Can you explain it?
Cheers
Describe explicitly all homomorphisms h: C_3 --->Aut(C_13). Hence describe all groups of order 39. How many such groups are there , up to isomorphism?
What do they mean by up to isomorphism?.Can you explain it?
Cheers
I presume you know that an isomorphism between groups G and H is an invertible function from G to H that "preserves the operation". If x, y are in G, the operation in G is denonted "*", and the operation in H is denonted "o", then f(x*y) must equal f(x)og(y). A few of the things you can show about isomorophisms are "identity maps to identity", "if x maps to y then $\displaystyle x^{-1}$ maps to $\displaystyle y^{-1}$", "if x maps to y and x has the property that $\displaystyle x^n$ equals the identity, the so does y".
Isomorphism, of course, is an equivalence relation. "How many up to isomorphism" is really asking "how many equivalence classes are there?" It is really asking "How many different groups are there?" where here "different" means having different properties.
In general if two groups are "isomorphic", they have the exactly same properties. One can be thought of as really the same group, just "relabled".
Here is a very simple version of your original question:
How many groups are there of order 2?
Well, $\displaystyle Z_2$ is such: it has 2 members, 0 and 1 and the operations are : 0+0= 0, 1+0= 1, 0+ 1= 1, 1+1= 0. Another is the "symmetries of a directed line segment". If [a,b] is the directed line segment (a line segement with a particular order: a to b, say, rather then b to a, where a and b are its endpoints), its "symmetries" are of order 2 because I can have the line segment itself, or the directd line segment [b, a], which is just [a, b] with reverse order. We can, as with symmetries of a square, think of [a, b] as the operation "don't change the directed line segment" and [b, a] as the operation "flip the line segment" and, with "+" meaning "do the two operations", [a, b]+ [a, b]= [a, b], [a, b]+ [b, a]= [b, a], [b,a]+ [a,b]= [b,a], [b, a]+[b,a]= [a, b]. That is a different group but it is easy to see that the mapping 0-> [a,b], 1-> [b,a] is an isomorphism.
If fact, given any two symbols, x, y, it is easy to define x+ x= x, x+ y= y, y+ x= y, y+ y= x and have a group of order 2. Since there are an infinite number of symbols, there are an infinite number of groups of any order.
But since an isomorphism must map the identity of to the identity of the other group, and there is only one element left in either group, map those to each other: x-> 0, y-> 1 and it is easy to show that is an isomorphism.
Although there are an infinite number of groups any order, all groups of order 2 are isomorphic so there is only one group of order 1 "up to isomorphism".
It is easy to prove that, if n is prime, there is only one group of order n "up to isomorphism". There are, however, two "different" groups of order 4: the rotation group and the "Klein 4 group". The Klein 4 group has the property that every member is its own inverse and the rotation group does not. But all groups of order 4 are isomorphic to one of those two so there are 2 groups of order 4 "up to isomorphism".
Since in order to be able to do the problem posted, you surely would have to know about "isomorphism" and other things, I suspect I have "over' answered your question!
In order to answer your question, you need to use the fact that 38= 2(19), a product of two prime numbers.