Let G = {1, -1, i, -i} be the multiplication group, and H = {1. -1} is a subgroup of G, then
Understanding the Group and Subgroup
We are given the group \(G = \{1, -1, i, -i\}\) under multiplication. This is a well-known group, often referred to as the group of 4th roots of unity. The identity element is 1. It's a cyclic group, generated by \(i\).
We are also given the subset \(H = \{1, -1\}\). We need to check if \(H\) is a normal subgroup of \(G\).
What is a Normal Subgroup?
In Group Theory, a subgroup \(H\) of a group \(G\) is called a normal subgroup if for every element \(g \in G\) and every element \(h \in H\), the element \(ghg^{-1}\) is in \(H\). This is often written as \(gHg^{-1} \subseteq H\) for all \(g \in G\). Since we are working with finite groups, this is equivalent to \(gHg^{-1} = H\) for all \(g \in G\).
An important property to remember is that every subgroup of an abelian group is a normal subgroup. Let's first check if \(G\) is abelian. For multiplication, \(ab = ba\) must hold for all \(a, b \in G\).
- \(1 \times (-1) = -1\), \(-1 \times 1 = -1\)
- \(i \times (-i) = -i^2 = -(-1) = 1\), \(-i \times i = -i^2 = 1\)
- \(i \times (-1) = -i\), \(-1 \times i = -i\)
- \(1 \times i = i\), \(i \times 1 = i\)
- and so on...
We can see that multiplication in \(G\) is commutative. For example, \(i \times (-1) = -i\) and \(-1 \times i = -i\). In general, for any \(a, b \in G\), \(ab = ba\). Thus, \(G\) is an abelian group.
Checking the Normal Subgroup Condition for Subgroup H
Since \(G\) is an abelian group, any subgroup of \(G\) is a normal subgroup. As \(H = \{1, -1\}\) is a subset of \(G\) and forms a group under multiplication (1x1=1, 1x-1=-1, -1x1=-1, -1x-1=1), \(H\) is indeed a subgroup of \(G\).
Therefore, based on the property of abelian groups, \(H\) must be a normal subgroup of \(G\).
Let's also verify this directly using the definition \(ghg^{-1} \in H\) for all \(g \in G\) and \(h \in H\). The elements of \(G\) are \(1, -1, i, -i\). The elements of \(H\) are \(1, -1\). The inverse of each element in \(G\) is: \(1^{-1}=1\), \((-1)^{-1}=-1\), \(i^{-1}=-i\), \((-i)^{-1}=i\).
We compute \(ghg^{-1}\) for all combinations:
| \(g \in G\) | \(h \in H\) | \(g^{-1}\) | \(ghg^{-1}\) Calculation | Result | Is Result in \(H\)? |
|---|---|---|---|---|---|
| 1 | 1 | 1 | \(1 \times 1 \times 1\) | 1 | Yes |
| 1 | -1 | 1 | \(1 \times (-1) \times 1\) | -1 | Yes |
| -1 | 1 | -1 | \((-1) \times 1 \times (-1)\) | 1 | Yes |
| -1 | -1 | -1 | \((-1) \times (-1) \times (-1) = 1 \times (-1)\) | -1 | Yes |
| i | 1 | -i | \(i \times 1 \times (-i)\) | \(i \times (-i) = -i^2 = 1\) | Yes |
| i | -1 | -i | \(i \times (-1) \times (-i)\) | \((-i) \times (-i) = i^2 = -1\) | Yes |
| -i | 1 | i | \((-i) \times 1 \times i\) | \((-i) \times i = -i^2 = 1\) | Yes |
| -i | -1 | i | \((-i) \times (-1) \times i\) | \(i \times i = i^2 = -1\) | Yes |
Conclusion on Subgroup H being Normal
As shown in the table, for every element \(g \in G\) and every element \(h \in H\), the element \(ghg^{-1}\) is either 1 or -1, both of which are elements of \(H = \{1, -1\}\). Thus, \(ghg^{-1} \in H\) for all \(g \in G\) and \(h \in H\).
This confirms that \(H\) is indeed a normal subgroup of \(G\).
Therefore, the statement "H is a normal subgroup of G" is correct.
Important Questions from Group & Subgroups
Name the smallest non cyclic group.
The generator of the group G = {a, a2, a3, a4, a5, a6 = e} is
Let H be a subgroup of a group G and K be a normal subgroup of a group G, then
A commutative group G is simple if and only if-
If G is a group such that a2 = e for all a ∈ G, then G is:
