7. If [G: H] = 2, prove that gH = Hg (18 from 6.5)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Help with number 7 please
![**Problem 7:**
**Statement:** If \([G : H] = 2\), prove that \( gH = Hg \).
*Reference:* (18 from 6.5)
**Explanation:**
This problem involves group theory concepts in abstract algebra. Here, \([G : H]\) represents the index of the subgroup \(H\) in the group \(G\), which is given as 2. The task is to prove the equality of the left coset \(gH\) and the right coset \(Hg\).
**Solution Outline:**
Given:
- \(G\) is a group, and \(H\) is a subgroup of \(G\).
- The index \([G : H] = 2\).
We need to show:
- \( gH = Hg \) for any \( g \in G \).
**Steps to Prove:**
1. Use the fact that \( |G| / |H| = 2 \), which implies that there are only two distinct cosets of \(H\) in \(G\).
2. Show that if \(Hg \neq H\), then \(Hg\) must be the other coset \(gH\), based on the property of groups and cosets with index 2.
3. Prove that for any \( g \in G \), the equality \( gH = Hg \) holds, implying that the left and right cosets are the same.
This type of problem is fundamental in understanding the structure of subgroups and cosets within group theory. The provided reference might lead to further theorems or properties in a textbook or lecture notes (18 from 6.5 suggests it might be Chapter or Section references within these notes).
**Conclusion:**
Upon following the steps, it can be concluded that when the index of \(H\) in \(G\) is 2, every element \(g\) in \(G\) forms a coset with \(H\) that is invariant whether taken to the left or right, hence proving \( gH = Hg \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fed14a3ea-da26-4be7-a143-8b845df95e91%2F6d5026d1-6be6-4cdf-837f-e2347bb597a4%2Frp3cefg_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 7:**
**Statement:** If \([G : H] = 2\), prove that \( gH = Hg \).
*Reference:* (18 from 6.5)
**Explanation:**
This problem involves group theory concepts in abstract algebra. Here, \([G : H]\) represents the index of the subgroup \(H\) in the group \(G\), which is given as 2. The task is to prove the equality of the left coset \(gH\) and the right coset \(Hg\).
**Solution Outline:**
Given:
- \(G\) is a group, and \(H\) is a subgroup of \(G\).
- The index \([G : H] = 2\).
We need to show:
- \( gH = Hg \) for any \( g \in G \).
**Steps to Prove:**
1. Use the fact that \( |G| / |H| = 2 \), which implies that there are only two distinct cosets of \(H\) in \(G\).
2. Show that if \(Hg \neq H\), then \(Hg\) must be the other coset \(gH\), based on the property of groups and cosets with index 2.
3. Prove that for any \( g \in G \), the equality \( gH = Hg \) holds, implying that the left and right cosets are the same.
This type of problem is fundamental in understanding the structure of subgroups and cosets within group theory. The provided reference might lead to further theorems or properties in a textbook or lecture notes (18 from 6.5 suggests it might be Chapter or Section references within these notes).
**Conclusion:**
Upon following the steps, it can be concluded that when the index of \(H\) in \(G\) is 2, every element \(g\) in \(G\) forms a coset with \(H\) that is invariant whether taken to the left or right, hence proving \( gH = Hg \).
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