If G is a group and n>= 1, define G(n) = {x in G : ord(x) = n}
If G =~ H ( isomorphism ), show that for all n>= 1, |G(n)| = |H(n)|
deduce that C3* C3 is not isomorphic C9 ( where C3, C9 are the cyclic groups of order 3 and 9 respectively)
I have no clue how to do this question. Please help
Is it true that C3 * C5 =~ C15? I think the answer is YES because 3 and 5 are prime numbers, so the groups of these order are cyclic. Am I right?
Is it true that C2 * C6 =~ C12? I think not maybe because 6 is not a prime number.
Thank you for your time, I'm really appreciated
basically its a property that if G =~ H ( isomorphism ) then |G(n)| = |H(n)|If G is a group and n>= 1, define G(n) = {x in G : ord(x) = n}
If G =~ H ( isomorphism ), show that for all n>= 1, |G(n)| = |H(n)|
deduce that C3* C3 is not isomorphic C9 ( where C3, C9 are the cyclic groups of order 3 and 9 respectively)
I have no clue how to do this question. Please help
Is it true that C3 * C5 =~ C15? I think the answer is YES because 3 and 5 are prime numbers, so the groups of these order are cyclic. Am I right?
Is it true that C2 * C6 =~ C12? I think not maybe because 6 is not a prime number.
Thank you for your time, I'm really appreciated
if you need a proof then do tell i wlll post it
Well, if two groups in general are isomorphic then by definition there exists a bijective mapping from one group to the other.
In order for us to have this mapping, we must have that both groups have the same number of elements (two sets have the same number of elements iff there exists a bijection between them).
Thus, the two groups must have the same order.
I wonder why you ask when, ThePerfectHacker has clearly explained in post #2 that:
All elements have the property that (Verify!). But needs has at least one element with order 12. ThusIs it true that C2 * C6 =~ C12? I think not maybe because 6 is not a prime number.
To Ursa, can you post me a proof of this problem please?
hi
well yes ThePerectHacker and o_O has clearly explained
i m jst converting in mathematical form
X Є G
and O(G)=n
=> X^n =e
=> f(X^n)=f(e)
=> f(X.X.X.X.X….X)=e’
=> f(X). f(X) . f(X) . f(X) . f(X)…… . f(X)=e’
=> (f(X))^n=e’
=> O(f(X))=n
=>O(H)=n
=>O(H)=O(G)
Since no one has answered the latter questions I will give you a hint. Look at Your intuition is right that this is isomorphic to . But why? Your hunch was because 3 and 5 were prime. But notice that because is not cyclic.
But what else do we notice about 3 and 5?