Linked Questions

2
votes
2answers
317 views

If $G$ is an infinite group, and $A,B$ subgroups of finite index in $ G$, then $A \cap B$ has finite index in $G$ [duplicate]

Question: If $G$ is an infinite group, and $A, B$ subgroups of finite index in $G$, then prove $A \cap B$ has finite index in $G$. I'm trying to show that $A\cap B$ can not have infinite index, but I ...
0
votes
1answer
109 views

A question on Cosets [duplicate]

Let $G$ be a group and $H$ , $K$ be subgroups of $G$ such that $[G:H]$ and $[G:K]$ are finite. Then is it true that $[G:H∩K]$ is also finite ?
0
votes
0answers
132 views

$[G : H] < ∞$ and$ [G : K] < ∞$ then $[G : H ∩ K] < ∞$ where $H$ and $K$ be subgroups of $G$ [duplicate]

Let $H$ and $K$ be subgroups of a group $G$. Then is the following true ? If $[G : H] < ∞$ and $[G : K] < ∞$, then $[G : H ∩ K] < ∞$. I think this's false because there's still a case ...
37
votes
5answers
20k views

Order of a product of subgroups. Prove that $o(HK) = \frac{o(H)o(K)}{o(H \cap K)}$.

Let $H$, $K$ be subgroups of $G$. Prove that $o(HK) = \frac{o(H)o(K)}{o(H \cap K)}$. I need this theorem to prove something.
28
votes
2answers
12k views

A group $G$ with a subgroup $H$ of index $n$ has a normal subgroup $K\subset H$ whose index in $G$ divides $n!$

I would be very thankful if someone could give me a hint with proving this. It's a very common exercise in abstract algebra textbooks. If $G$ is a group with a subgroup $H$ of finite index $n$, ...
14
votes
2answers
505 views

When are finite-index subgroups of a Lie group closed?

Let $G$ be a Lie group (or, if necessary, a reductive Lie group) and $H$ a subgroup of $G$. If $\lbrack G:H\rbrack < \infty$, is it true that $H$ is closed? If not, are there any broad assumptions ...
3
votes
3answers
1k views

prove that $H\cap K$ have finite index in G

If $G$ is a group and $H,K$ are two subgroups of finite index in $G$, prove that $H\cap K$ is of finite index in $G$. Can you find upper bound for index of $H\cap K$ in $G$? Since $a(H\cap K$) $\...
3
votes
3answers
947 views

$\left[G:H\cap K\right]=\left[G:H\right]\left[G:K\right]$ if $\left[G:H\right],\left[G:K\right]$ are coprime

Let $G$ be a finite group and $H<G, K<G$. I have shown that $\left[G:H\cap K\right]\leq\left[G:H\right]\left[G:K\right]$ But I do not know where to begin to prove the equality in case these ...
4
votes
1answer
3k views

Prove that intersection of finite index subgroups has finite index.

I'm trying this problem from Herstein: Q) If G is a group and H, K are two subgroups of finite index in G, prove that H $\cap$ K is of finite index in G. Can you find an upper bound for the index ...
1
vote
1answer
2k views

My approach to "Does the intersection of two finite index subgroups have finite index? [Please check my method]

This question/thread is continuation from here Also I.N Herstein in Page 40 Question 6 has brought up the same question My approach: We know the property $\left | G:K \right |\leq \left | G:H \right|\...
2
votes
2answers
160 views

If $[G:H]<\infty$, then $H$ contains a normal subgroup $N$ of $G$ such that $[G:N]<\infty$.

Let $G$ be a group and $H$ be a subgroup of $G$. I want to prove that if $[G:H]<\infty$, then $H$ contains a normal subgroup $N$ of $G$ such that $[G:N]<\infty$. Professor gave me the following ...
2
votes
1answer
379 views

Bounding the index of subgroup intersection

Full disclosure: this is a homework problem, but it is not assigned to turn in for credit. The problem is from Dummit and Foote, Chapter 3.2: Suppose $H, K$ are subgroups of finite index a group (not ...
1
vote
1answer
210 views

Index of intersection of finite index subgroups

I'm trying to understand this proof https://math.stackexchange.com/a/128549/205193 which does not seem complicated but I don't understand why : why $p(x)=p(y)$, implies that $x$ and $y$ are in the ...
0
votes
1answer
86 views

On formula $\sum_{i=1}^n 1/(G : H_i) = 1$ on a group $G$

Let $G$ be a group. Let $H$ be a subgroup of $G$ such that $(G : H) \lt \infty$. Then there exists a sequence of elements $a_1,\cdots, a_n$ such that $G = \bigcup_{i=1}^n a_iH$ is a disjoint union. ...
0
votes
1answer
93 views

If $H$ and $K$ are subgroups of $G$, what does the notation $HK$ mean?

I know there is direct product $H \times K$ and semidirect product, but what is implied when it is just $HK$ without any symbols in between? For example, in this question it says $[HK:K] \leq [G:K]$. ...