13,384 reputation
12062
bio website
location Everywhere
age
visits member for 3 years, 3 months
seen 19 mins ago

.


May
29
comment find a group with the property :
This subgroup is the direct sum of the $\mathbb{Z}_n$, and is often denoted by $\mathbb{Z}_1 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \ldots$
May
27
awarded  Popular Question
May
21
comment Schur's Lemma in Group Theory
Also, $G = \operatorname{SL}_2(F)$ is an example for any finite field $F$ other than $\mathbb{F}_2$ and $\mathbb{F}_3$. The reason is the same, the only nontrivial proper subgroup of $G$ is $Z(G) = \{I, -I\}$.
May
19
comment Extending a homomorphism $f:\left<a \right>\to\Bbb T$ to $g:G\to \Bbb T$, where $G$ is abelian and $\mathbb{T}$ is the circle group.
Just like before, $H = \oplus_{a \in A} \mathbb{Z}$, the direct sum of $|A|$ copies of $\mathbb{Z}$, ie. the free abelian group with basis $A$.
May
15
comment Extending a homomorphism $f:\left<a \right>\to\Bbb T$ to $g:G\to \Bbb T$, where $G$ is abelian and $\mathbb{T}$ is the circle group.
@AlexB.: Let $H = \oplus_{a \in A} \mathbb{Z}$, ie. the free abelian group with basis $A$. Then there exists a surjective homomorphism $H \rightarrow A$. Also, we can choose $D = \oplus_{a \in A} \mathbb{Q}$.
May
9
answered Quickest route to the structure theorem for finitely generated modules over a PID
May
8
comment Invariants of a subgroup.
As in the book, define $G(r) = \{ x \in G: x^r = 1\}$. To prove the first inequality $k \geq s$, you can use the corollary of lemma 2.14.2, which states that $|G(p)| = p^k$.
May
7
comment $D_8$ as a derived subgroup
Related: math.stackexchange.com/questions/361582/… and math.stackexchange.com/questions/365679/…
May
6
awarded  Caucus
May
4
answered Extending a homomorphism $f:\left<a \right>\to\Bbb T$ to $g:G\to \Bbb T$, where $G$ is abelian and $\mathbb{T}$ is the circle group.
May
4
comment Extending a homomorphism $f:\left<a \right>\to\Bbb T$ to $g:G\to \Bbb T$, where $G$ is abelian and $\mathbb{T}$ is the circle group.
@AlexanderGruber: How about this? Let $A$ be an abelian group such that for any abelian group $G$ and subgroup $H \leq G$, any homomorphism $f: H \rightarrow A$ can be extended to a homomorphism $g: G \rightarrow A$. Now consider the identity map $f: A \rightarrow A$. We can embed $A$ (any abelian group, in fact) into a divisible group $D$, so we can extend $f$ to a surjective homomorphism $g: D \rightarrow A$. Quotients of divisible groups are divisible, so $A$ is divisible. Thus for abelian groups, this type of homomorphism extension is possible if and only if the target group is divisible.
May
4
comment Extending a homomorphism $f:\left<a \right>\to\Bbb T$ to $g:G\to \Bbb T$, where $G$ is abelian and $\mathbb{T}$ is the circle group.
I think this is true, and even more generally when $\langle a \rangle$ is replaced by any subgroup of $G$ and $\mathbb{T}$ is replaced by any divisible abelian group.
May
4
comment Normal subgroup of a normal subgroup
Sure, but it's nice to see a general example.
May
4
answered Normal subgroup of a normal subgroup
May
4
comment Normal subgroup of a normal subgroup
@user75908: $G$ contains every permutation with the cycle structure $(ab)(cd)$ and conjugation by a permutation preserves cycle structure.
May
3
answered Terminology for infinite groups, all of whose subgroup have finite index.
May
2
comment Terminology for infinite groups, all of whose subgroup have finite index.
I think the only such group is $\mathbb{Z}$. If every nontrivial subgroup of $G$ is of finite index and $G$ is infinite, then $G$ is torsionfree and contains a subgroup of finite index isomorphic to $\mathbb{Z}$. The only group with this property is $\mathbb{Z}$.
Apr
30
comment Proper subgroup of $\mathbb{Q}^{+}$ with finite index
@PeterTamaroff: It was before the edit, when this question contained five different questions, and the only relation between them was group theory. I don't think questions like that belong here. Now the question is okay, but it seems that removing close votes is not possible
Apr
30
answered $\bigcup_{x \in G} xHx^{-1} \neq G$
Apr
30
answered There is an automorphism of $\mathbb Z_6$ which is not an inner automorphism