Let G be an abelian group. prove: H = {g € G||g| <∞o} is a subgroup of G. We need to prove that H ‡ Ø) and Vx, y ≤ H : xy¯¹ ɛ H. We know that e € H, because |e| = 1 < ∞, H‡Ø. Furthermore, if we take x, y = H, we know that xª = yb=1 for a, b = Z, so (xy−¹)ab = (xª)b(yb)−ª = 1, -1 › |xy¯¹| < ∞ and xy EH. G is not abelian, H doesn't have to be a subgroup. For example choose G = D∞, so H = {d€ Do | |d| <∞}. oth sr and sr² have order 2, but srsr² = r has infinite order, so r H. Therefore H is not a bgroup of G.
Let G be an abelian group. prove: H = {g € G||g| <∞o} is a subgroup of G. We need to prove that H ‡ Ø) and Vx, y ≤ H : xy¯¹ ɛ H. We know that e € H, because |e| = 1 < ∞, H‡Ø. Furthermore, if we take x, y = H, we know that xª = yb=1 for a, b = Z, so (xy−¹)ab = (xª)b(yb)−ª = 1, -1 › |xy¯¹| < ∞ and xy EH. G is not abelian, H doesn't have to be a subgroup. For example choose G = D∞, so H = {d€ Do | |d| <∞}. oth sr and sr² have order 2, but srsr² = r has infinite order, so r H. Therefore H is not a bgroup of G.
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter4: More On Groups
Section4.4: Cosets Of A Subgroup
Problem 30E: Let G be an abelian group of order 2n, where n is odd. Use Lagranges Theorem to prove that G...
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I have the following question and solution, the assignment is to give feedback about the solution and the way its written.
![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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F42044228-cff6-4391-9596-005bad4f5c7f%2F7c92dbc4-63a5-425f-b453-b238379d60ee%2Fetjti1p_processed.png&w=3840&q=75)
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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F42044228-cff6-4391-9596-005bad4f5c7f%2F7c92dbc4-63a5-425f-b453-b238379d60ee%2Fl7aywo7_processed.png&w=3840&q=75)
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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