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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
Apr
27
comment Reference of a Theorem in Group Theory
@PeterTamaroff: Something like "A transitive group $G$ on $m$ letters $a, b, c, \ldots$ contains necessarily a permutation that moves every letter"
Apr
27
answered Reference of a Theorem in Group Theory
Apr
26
comment When is the group of units in $\mathbb{Z}_n$ cyclic?
Groups are always nonempty since they contain the identity.
Apr
26
revised If N is a normal subgroup of G,decide whether np(N) | np(G) or np(G/N) | np(G)?
edited body
Apr
26
revised When is the group of units in $\mathbb{Z}_n$ cyclic?
added 697 characters in body
Apr
26
answered When is the group of units in $\mathbb{Z}_n$ cyclic?
Apr
26
answered If N is a normal subgroup of G,decide whether np(N) | np(G) or np(G/N) | np(G)?
Apr
25
revised Number of Homomorphisms from $\Bbb{Z}_m$ to $\Bbb{Z}_n$
edited title
Apr
24
comment Groups with 20 Sylow subgroups
@SteveD: That would simplify the proof. How do you find the subgroup of index $5$ or $19$ when $n_2 = 95$?
Apr
24
comment Proving that a group of order $112$ is not simple
@Ziggy: If $G$ is a simple group and $[G:H] = n$, then the left coset action gives us an injective homomorphism $f: G \rightarrow S_n$ (in this case $H = N_G(P)$ for $P$ some Sylow $2$-subgroup). With Sylow subgroups you could apply the conjugation action as well.
Apr
23
answered Groups with 20 Sylow subgroups
Apr
23
comment Groups with 20 Sylow subgroups
How do you reduce to cases (1), (2) and (3)?
Apr
21
comment Let $G$ be a simple group such that $|G|=p^2q r$ ($p$ and $q$ are distinct prime numbers) then how prove $|G|=60$?
Ah, I didn't notice that part about $A_r$ in the first bullet point.
Apr
21
comment Let $G$ be a simple group such that $|G|=p^2q r$ ($p$ and $q$ are distinct prime numbers) then how prove $|G|=60$?
With BTT we can prove that if a Sylow subgroup corresponding to the smallest prime divisor is cyclic, then it has a normal complement. So that immediately implies that $p$ is the smallest prime divisor and we can assume $p < q < r$. And to prove $G \cong A_5$, we can use BTT again (it implies that $G$ has exactly $5$ Sylow $2$-subgroups).
Apr
21
comment A group of order 105 contains a subgroup of order 35.
Note that it is true that every group of order $105$ will have a normal Sylow $7$-subgroup (because the subgroup of order $35$ is normal and contains a unique subgroup of order $7$). See my answer
Apr
21
answered A group of order 105 contains a subgroup of order 35.