**Proof: Intersection of Two Subgroups****Statement:** Prove that the intersection of two subgroups of $ G $ is itself a subgroup of $ G $.**Proof Outline:**To prove this, we need to show that the intersection of two subgroups, say $ H $ and $ K $, of a group $ G $ is also a subgroup of $ G $. Let's denote the intersection by $ H \cap K $.**1. Non-Empty Intersection:** - Since both $ H $ and $ K $ are subgroups, they both contain the identity element $ e $ of $ G $. - Therefore, the identity element $ e $ is also in $ H \cap K $.**2. Closure:** - Take any two elements $ a, b \in H \cap K $. - Since $ a $ and $ b $ are in both $ H $ and $ K $, and both $ H $ and $ K $ are subgroups, the product $ ab $ is in $ H $ and $ ab $ is in $ K $. - Thus, $ ab $ is in $ H \cap K $.**3. Inverses:** - Take any element $ a \in H \cap K $. - Since $ a $ is in both $ H $ and $ K $, and both $ H $ and $ K $ are subgroups, the inverse $ a^{-1} $ is in $ H $ and $ a^{-1} $ is in $ K $. - Thus, $ a^{-1} $ is in $ H \cap K $.By satisfying these conditions (non-empty intersection, closure, and inverses), $ H \cap K $ is a subgroup of $ G $. This completes the proof.

Suppose that, H1 and H2 be the sub groups of G. Also, let H=H1∩H2 . Consider one step group test:…

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Prove that the intersection of two subgroups of G is itself a subgroup of G.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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**Proof: Intersection of Two Subgroups**

**Statement:** Prove that the intersection of two subgroups of $ G $ is itself a subgroup of $ G $.

**Proof Outline:**

To prove this, we need to show that the intersection of two subgroups, say $ H $ and $ K $, of a group $ G $ is also a subgroup of $ G $. Let's denote the intersection by $ H \cap K $.

**1. Non-Empty Intersection:**
- Since both $ H $ and $ K $ are subgroups, they both contain the identity element $ e $ of $ G $.
- Therefore, the identity element $ e $ is also in $ H \cap K $.

**2. Closure:**
- Take any two elements $ a, b \in H \cap K $.
- Since $ a $ and $ b $ are in both $ H $ and $ K $, and both $ H $ and $ K $ are subgroups, the product $ ab $ is in $ H $ and $ ab $ is in $ K $.
- Thus, $ ab $ is in $ H \cap K $.

**3. Inverses:**
- Take any element $ a \in H \cap K $.
- Since $ a $ is in both $ H $ and $ K $, and both $ H $ and $ K $ are subgroups, the inverse $ a^{-1} $ is in $ H $ and $ a^{-1} $ is in $ K $.
- Thus, $ a^{-1} $ is in $ H \cap K $.

By satisfying these conditions (non-empty intersection, closure, and inverses), $ H \cap K $ is a subgroup of $ G $.

This completes the proof.

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