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For questions about the study of algebraic structures consisting of a set of elements together with a well-defined binary operation that satisfies three conditions: associativity, identity and invertibility.
1
vote
1
answer
55
views
Finding the charctristics subgroup of $(\mathbb Q,+)$
I faced to the following problem and could to verify the first part of it:
Let $q\in\mathbb Q-\{0\}$ and let $v_q:(\mathbb Q,+)\to (\mathbb Q,+)$ is defined as $$v_q(t)=qt$$ Then proved that $v_q$ …
6
votes
2
answers
642
views
If $N\lhd H×K$ then $N$ is abelian or $N$ intersects one of $H$ or $K$ nontrivially
I am thinking on this problem:
If $N\lhd H×K$ then either $N$ is abelian or $N$ intersects one of $H$ or $K$ nontrivially.
I assume; $N$ is not abelian so, there is $(n,n')$ and $(m,m')$ in $N$ …
0
votes
1
answer
124
views
Is this true about a permutation group?
There was a question here on the site that was asked:
Let $G$ be a finitely generated abelian group. Then prove that it is not isomorphic to $G/N$, for every subgroup $N≠⟨1⟩$.
Would the answer for a …
1
vote
2
answers
215
views
$|G|=12$ and no elements of order $2$ in $Z(G)$
I am thinking on the following problem:
If $|G|=12$ and there is no element of order $2$ in its center then $3$-Sylow subgroup of $G$ cannot be normal in $G$.
I was told that to assume the $3$-S …
8
votes
2
answers
127
views
$|G-H|<\infty$ so $|G|<\infty$
Let $G$ is a group and $H<G$ such that $|G-H|<\infty$. Prove that $|G|<\infty$.
Truthfully, there is a hint for it:
$H$ cannot be an infinite subgroup.
It is clear if $|H|<\infty$, since $| …
2
votes
1
answer
106
views
Evaluating $H_G$ when $H\leq G$ is subgroup of all diagonal invertible matrices.
I am working on some basic context in Group Theory. I face to the following problem:
Let $G=GL(2,\Bbb Q)$ and $H\leq G$ be the subgroup of all diagonal invertible matrices. Find $$H_G=\cap_{g\in G …
0
votes
1
answer
182
views
Searching for a $1/2$ -transitive group
As a definition, if for a group $(G$|$\Omega)$; the orders of $G_{\omega}$ ($\omega$ in $\Omega$) are equal to eachother, then $G$ is said to be a $1/2$ -transitive group . Any example for such these …
2
votes
1
answer
139
views
Why does exactly $G'$ become a subgroup of a given group $G$?
Reviewing the book "An Introduction to the Group Theory" By J.J.Rotman, I have found out that I have neglected the way $G'$, the commutator subgroup, becomes a subgroup of a given group $G$.
Unfortun …
4
votes
1
answer
201
views
Hom$(G,A)$ $\cong$Hom $(G/G',A)$
Let $G$ is a group and $\varphi: G\longrightarrow A$ be any group homomorphism wherein $A$ is abelian. Then $$\text{Hom}(G,A)\cong \text{Hom} \left(\frac{G}{G'},A\right)$$
What we have here is, f …
3
votes
2
answers
91
views
Verifying that $\mathbb Q=\bigcup_{n\ge 1} H_n$
Let $G=(\mathbb Q,+)$ and $r_1=p_1/q_1, r_2=p_2/q_2\in G$. I want to prove that:
$\langle r_1,r_2\rangle\subseteq \langle\frac{1}{q_1q_2}\rangle$
If $r_1,r_2,...,r_n\in G$ then $\langle r_1,r_2,... …
3
votes
0
answers
98
views
Are $\bar{G_i}(\cong G_i)$ in $\prod_{i=1}^n G_i$ unique?
We know that:
Theorem: if $G=\prod_{i=1}^n G_i$ be the direct product of groups $G_1,G_2,...,G_n$ then there exists normal subgroups $\bar{G_i}\cong G_i( i=1...n)$ of $G$ such that $$G=\bar{G_1}\b …
2
votes
2
answers
737
views
Every polycyclic group has a normal poly-infinite cyclic subgroup of finite index
There is a theorem that states:
"Every polycyclic group has a normal poly-infinite cyclic subgroup of finite
index. "
I just read the proof of it and honestly found some difficulties in it. The main …
1
vote
1
answer
61
views
Is $K=K∩G_\alpha N=G_\alpha$ an error in this context?
There is a problem in Problems in Group Theory by J.D.Dixon :
2.51 If a permutation group $G$ contains a minimal normal subgroup $N$ which is both transitive and abelian, then $G$ is primitive.
…
1
vote
1
answer
63
views
Order of the main set $A$ is greater than 10?
A group $G$, acting on a set $A$, is a permutation group such that it has an element of order $30$. Prove that order of $A$ is greater than $10$.
Here, I supposed that, $|A|=6$ and wanted to make …
5
votes
3
answers
172
views
($\mathbb Q$,+) and $\mathbf{Z}_{p^\infty}$ has this property?
There is an exercise telling that every finite subset of group ($\mathbb Q$,+) or of group $\mathbf{Z}_{p^\infty}$ generates a cyclic group itself.
For the first group if $X= \left\{\frac{p_{1}}{q_{1 …