1. ## group isomorphism

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)

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

2. Originally Posted by knguyen2005
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)
If $\phi :G\to H$ is an isomorphism then the order of $x$ is same as order of $\phi(x)$. Thus, define a mapping from $G(n)$ to $H(n)$ by $a\mapsto \phi(a)$ to show $|G(n)| = |H(n)|$.

Now to show $C_3\times C_3\not \simeq C_9$ show that $|C_3\times C_3(3) |\not = |C_9 (3)|$.

3. 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)

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 you need a proof then do tell i wlll post it

4. ## To Ursa

Originally Posted by ursa
basically its a property that if G =~ H ( isomorphism ) then |G(n)| = |H(n)|
if you need a proof then do tell i wlll post it
To Ursa, can you post me a proof of this problem please?

5. 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.

6. Originally Posted by knguyen2005
To Ursa, can you post me a proof of this problem please?
I wonder why you ask when, ThePerfectHacker has clearly explained in post #2 that:

Originally Posted by ThePerfectHacker
If $\phi :G\to H$ is an isomorphism then the order of $x$ is same as order of $\phi(x)$. Thus, define a mapping from $G(n)$ to $H(n)$ by $a\mapsto \phi(a)$ to show $|G(n)| = |H(n)|$.

Now to show $C_3\times C_3\not \simeq C_9$ show that $|C_3\times C_3(3) |\not = |C_9 (3)|$.
Is it true that C2 * C6 =~ C12? I think not maybe because 6 is not a prime number.
All elements $x \in C_2 \times C_6$ have the property that $x^6 = 1$ (Verify!). But $C_{12}$ needs has at least one element with order 12. Thus $C_{12} \not \cong C_2 \times C_6$

7. 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)

8. Since no one has answered the latter questions I will give you a hint. Look at $C_3 \times C_5$ Your intuition is right that this is isomorphic to $C_{15}$. But why? Your hunch was because 3 and 5 were prime. But notice that $C_3 \times C_3 \not \simeq C_9$ because $C_3 \times C_3$ is not cyclic.

But what else do we notice about 3 and 5?

9. I think its important to point out whats going on behind the scenes of gropus of the form $C_m\times{C_n}$. Maybe proving or disproving this claim can give you some insight.

Claim: $C_m\times{C_n}\simeq{C_{mn}}$ if and only if $gcd(m,n)=1$