Problem 3
Question
Let \(G\) be a group, and \(H\) a normal subgroup of \(G\). Prove the following: If \((G: H)=p\), where \(p\) is a prime, then the order of every element \(a \notin H\) in \(G\) is a multiple of \(p\). [Use (1).]
Step-by-Step Solution
Verified Answer
If the index of \(H\) in \(G\) is a prime \(p\), then any element in \(G\) but not in \(H\) has order that is a multiple of \(p\).
1Step 1: Understanding the problem
We need to prove that if the index of a normal subgroup \(H\) in the group \(G\) is a prime \(p\), then the order of any element \(a\) in \(G\) but not in \(H\) must be a multiple of \(p\). The index \((G: H) = p\) means there are \(p\) distinct cosets of \(H\) in \(G\).
2Step 2: Applying the property of index
Since \(H\) is normal in \(G\), the cosets form a group \(G/H\) where the number of elements is \(p\). \(G/H\), therefore, is a group of prime order and is cyclic. Any non-identity element in \(G/H\) must generate the entire group and thus have order \(p\).
3Step 3: Mapping elements to cosets
Select an element \(a \in G\) such that \(a otin H\). In \(G/H\), this translates to the coset \(aH\) not being the identity element coset \(H\). Since \(G/H\) is cyclic of prime order \(p\), the element \(aH\) must have order \(p\) in \(G/H\).
4Step 4: Relating the order in \(G/H\) to the order in \(G\)
If the coset \(aH\) in \(G/H\) has order \(p\), it implies that \((aH)^p = H\). This translates back to \(a^p \in H\). Because \(p\) is the smallest positive integer for which this is true, the order of \(a\) must be \(kp\) for some integer \(k\). Thus, the order of \(a\) in \(G\) is a multiple of \(p\).
5Step 5: Conclusion
We have shown through mapping to the quotient group and using its properties that if \(a otin H\), then the order of \(a\) is \(kp\). Therefore, every element \(a otin H\) in \(G\) has an order that is a multiple of \(p\), completing our proof.
Key Concepts
Normal SubgroupQuotient GroupCyclic GroupPrime OrderGroup Index
Normal Subgroup
A normal subgroup is a special type of subgroup where the group behaves in a very symmetric way with respect to its elements. For any element in the group and any element in this subgroup, if you conjugate the subgroup's element using the group's element, it will stay within the subgroup. In simpler terms, no matter how you mix elements from the subgroup with those from the group, using conjugation, they always end up back within the subgroup. Being normal is important because it allows the construction of a new structure called a Quotient Group.
Quotient Group
The Quotient Group is the collection of cosets of a normal subgroup in a group. When a subgroup is normal, all of its cosets can be grouped into a new group called the quotient group, denoted as \( G/H \). This structure is very interesting because it essentially takes the original group and 'simplifies' it into a new group, capturing some vital features of the original group's structure. The quotient group inherits some properties from the original group but in a compressed form. With a prime order quotient group, as in our exercise, the group is always cyclic.
Cyclic Group
In group theory, a cyclic group is a group that can be generated by a single element. This means that every group member can be expressed as some power of this single element. If you can think of spinning around a clock, and each tick being reachable by starting at zero and making repeated increments, that's a visual way to imagine it. One of the neat things about cyclic groups is that they have a very predictable, round structure, with elements cycling back to the start after a certain point. The quotient group formed when a normal subgroup's index in its group is prime, is cyclic.
Prime Order
A group having prime order means it contains a number of elements equal to some prime number \( p \). This is special because such groups have a strong and simple structure. If the group order is a prime number, the only possible subgroups it can have are the trivial sub-group containing just the identity and the group itself. Furthermore, all non-identity elements in a prime order group must generate the entire group. This simplifies the arithmetic within these groups to a great degree and reveals a tidy cyclic pattern.
Group Index
The index of a subgroup in a group is a count of how many times the subgroup can "fit" inside the group, distinct from each fit being seen as a coset. To picture this, think of reshuffling books into distinct pyramids where each pyramid represents a coset. The index is then the number of these pyramids, showing how much larger the group is compared to the subgroup. In the discussed exercise, a prime index indicates that the larger group can only be broken into a prime number of such pyramids, making it highly systematic, and important properties stemming from the prime nature apply.
Other exercises in this chapter
Problem 2
In each of the following, \(H\) is a subset of \(\mathbb{R} \times \mathbb{R}\) (a) Prove that \(H\) is a normal subgroup of \(\mathbb{R} \times \mathbb{R}\). (
View solution Problem 2
In each of the following, \(G\) is a group and \(H\) is a normal subgroup of \(G\). List the elements of \(G / H\) and then write the table of \(G / H\). Exampl
View solution Problem 3
There are some group properties which, if they are true in \(G / H\) and in \(H\), must be true in \(G .\) Here is a sampling. Let \(G\) be a group, and \(H\) a
View solution Problem 4
The center of a group \(G\) is the normal subgroup \(C\) of \(G\) consisting of all those elements of \(G\) which commute with every element of \(G\). Suppose t
View solution