I have two questions... I would really appreciate some help...
1) Suppose phi is a homomorphism from G to G' and that G' has an element of order 8. Prove that G has an element of order 8.
(I can show that G has an element g^8 that maps to the identity in G', but I think this shows only that g^8 is in the kernal of G... I'm stuck on how to show that it has an element of order 8)
2) Suppose G is a cyclic group and that 6 divides the order of G. How many elements of order 6 does G have?
(I have no idea what to do with this... wouldn't it depend on how large G is?)
any help or hints are very appreciated! thanks!
Sure, I can help you with both questions.
1) To prove that G has an element of order 8, given that G' has an element of order 8, we can use the fact that homomorphisms preserve order. Let's call the element of order 8 in G' as g', and its pre-image in G as g. Since phi is a homomorphism, we have phi(g) = g'.
Now, let's consider the order of g in G. Suppose the order of g is k, where k is a positive integer. We know that g^k = e, where e is the identity element in G. Applying the homomorphism phi to both sides, we get phi(g^k) = phi(e). Since phi is a homomorphism, we can rewrite this as (phi(g))^k = e'.
Since g' has an order of 8 in G', and phi(g) = g', we know that (phi(g))^8 = e'. Therefore, k must be a multiple of 8 to satisfy the equation (phi(g))^k = e'. Hence, G must have an element of order 8.
2) If G is a cyclic group and 6 divides the order of G, we can use Lagrange's theorem which states that the order of any subgroup of a finite group divides the order of the group.
Since G is cyclic, it has a generator, and every element in G can be expressed as a power of this generator. Let's call the generator of G as g. Suppose the order of G is n, and we know that 6 divides n.
Now, let's consider the elements of G that have order 6. First, note that an element g^k has order 6 if and only if the greatest common divisor of k and n is 6. In other words, gcd(k, n) = 6.
To count the number of elements of order 6, we need to count the number of integers k between 1 and n (inclusive) such that gcd(k, n) = 6. This can be done by counting the number of positive integers that divide both 6 and n, which is equivalent to counting the number of positive divisors of n that are divisors of 6 (since all divisors of n are positive).
Therefore, to find the number of elements of order 6 in G, we need to find the number of positive divisors of n that are divisors of 6.
Sure, I can help you with both questions!
1) To prove that G has an element of order 8, we can use the given information that phi is a homomorphism from G to G' and that G' has an element of order 8.
Since G' has an element of order 8, let's call that element a' in G'. We want to find an element a in G such that phi(a) = a'. One way to do this is to find an element b in G such that phi(b^8) = a'.
Let's consider the element g = b^8 in G. Since phi is a homomorphism, we have phi(g) = phi(b)^8. We know that phi(g) = a' and phi(b)^8 = (phi(b))^8 (since phi is a homomorphism). Therefore, we have a' = (phi(b))^8.
Since a' has order 8, we know that (a')^8 = e', where e' is the identity element in G'. So, we can write (phi(b))^8 = e'. This implies that phi(b) has order 8 in G'.
Now, we need to show that b has order 8 in G. Suppose b has order k in G. Since phi is a homomorphism, we have (phi(b))^k = phi(b^k). Since phi(b) has order 8 in G', we know that (phi(b))^8 = e'. So, we have e' = (phi(b))^k = phi(b^k).
Since the homomorphism phi maps the identity element of G to the identity element of G', we know that phi(e) = e'. Therefore, we have e' = phi(b^k) = phi(e). This implies that b^k = e.
We started assuming that b has order k in G, but now we know that b^k = e. Therefore, k divides 8. Since 8 is not divisible by any number less than 8 (since 8 is prime), we conclude that k = 8, and hence, b has order 8 in G.
Therefore, G must have an element of order 8.
2) If G is a cyclic group and 6 divides the order of G, then let's say the order of G is n. We know that the order of an element a in a cyclic group is equal to the order of the group itself when a is a generator.
Since G is cyclic, it must have a generator, let's call it g. We want to find how many elements of order 6 G has.
Now, since g generates G and the order of G is n, we know that g^n = e, where e is the identity element of G. We also know that the order of g is n.
Since 6 divides n, we can write n = 6k for some positive integer k.
Now, let's consider the element h = g^k. Since the order of g is n, we have (g^k)^n = g^(k * n) = g^(6k) = e. This means that the order of h is either 1, 2, 3, or 6.
Let's analyze these cases:
- If the order of h is 1, this means that h = e, the identity element. But we can see that g^k = e only if k is a multiple of n, which contradicts our assumption that k is a positive integer.
- If the order of h is 2, this means that h^2 = e. But we can see that (g^k)^2 = e only if k is a multiple of 3 (since (g^k)^2 = g^(2k) = e), which contradicts our assumption that k is a positive integer.
- If the order of h is 3, this means that h^3 = e. But we can see that (g^k)^3 = e only if k is a multiple of 2 (since (g^k)^3 = g^(3k) = e), which contradicts our assumption that k is a positive integer.
- If the order of h is 6, this means that h^6 = e. This is true if k is any positive integer.
Therefore, if G is a cyclic group and 6 divides the order of G, then G has exactly one element of order 6.
I hope this explanation helps! If you have any further questions, please let me know.