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Let G be an abelian group. Prove that (g = G||g|

Let G be an abelian group. Prove that (g = G||g|

Advanced Engineering Mathematics
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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The following question and solution are in the attached images, show what could've been written better and give feedback on the written proof, thank you in advance

Let G be an abelian group. To prove: H = {g €G ||g| <∞} is a subgroup of G. We need to prove that H ‡ Ø and Vx, y ≤H : xy-¹ € H. We know that e € H, because |e| = 1 < ∞, so H ‡ Ø). Furthermore, if we take x, y € H, we know that xª = y³ = 1 for a, b = Z, so -a (xy-¹)ab = (xª)b(y³)¯ = 1, so xy-¹ <∞ and xy-¹ € H. If G is not abelian, H doesn't have to be a subgroup. For example choose G = D∞, so H = {de D∞ | |d| <∞}. Both sr and sr² have order 2, but srsr² = r has infinite order, so r H. Therefore H is not a subgroup of G.
Transcribed Image Text:Let G be an abelian group. To prove: H = {g €G ||g| <∞} is a subgroup of G. We need to prove that H ‡ Ø and Vx, y ≤H : xy-¹ € H. We know that e € H, because |e| = 1 < ∞, so H ‡ Ø). Furthermore, if we take x, y € H, we know that xª = y³ = 1 for a, b = Z, so -a (xy-¹)ab = (xª)b(y³)¯ = 1, so xy-¹ <∞ and xy-¹ € H. If G is not abelian, H doesn't have to be a subgroup. For example choose G = D∞, so H = {de D∞ | |d| <∞}. Both sr and sr² have order 2, but srsr² = r has infinite order, so r H. Therefore H is not a subgroup of G.
Let G be an abelian group. Prove that {g G||g| < oo}) is a subgroup of G (called the torsion subgroup of G). Give an explicit example where this set is not a subgroup when G is non-abelian.
Transcribed Image Text:Let G be an abelian group. Prove that {g G||g| < oo}) is a subgroup of G (called the torsion subgroup of G). Give an explicit example where this set is not a subgroup when G is non-abelian.
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