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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.
Apr
21
comment When can I say that $\cup H_i$ is a group?
The sufficent condition you mention says that the poset of subgroups (ordered by inclusion) is a directed set. en.wikipedia.org/wiki/Directed_set
Apr
21
comment When can I say that $\cup H_i$ is a group?
More generally, the union $\cup H_i$ is a subgroup when for all $i, j \in I$ there exists $k \in I$ such that $H_i \leq H_k$ and $H_j \leq H_k$. This is not a necessary condition as the Klein $4$-group shows.
Apr
21
comment When can I say that $\cup H_i$ is a group?
The problem with your generalization is that for example, when $H_1 \cup H_2 \cup H_3$ is a subgroup, it does not necessarily follow that $H_2 \cup H_3$ is a subgroup.
Apr
21
comment When can I say that $\cup H_i$ is a group?
What kind of generalization did you have in mind?
Apr
20
comment Solve the quadratic congrunce
You can divide by two modulo $11$, so you can complete the square. The same idea should be helpful for solving quadratics $x^2 + bx + c \equiv 0 \mod{p}$ when $p$ is an odd prime.
Apr
20
answered $G$ is a nonabelian finite group, then $|Z(G)|\leq \frac{1}{4}|G|$
Apr
19
comment on a group with perfect automorphism group
What you define here as a "perfect group" is called a complete group (wikipedia). A group is called perfect when it equals its commutator subgroup (mathworld). These are two different things.
Apr
18
revised On the commutator subgroup of a group
deleted 20 characters in body
Apr
18
comment Which groups are derived subgroups?
Yet another reason: If $G'/G''$ and $G''$ are cyclic, then $G'' = 1$. See here
Apr
18
answered On the commutator subgroup of a group
Apr
16
comment Nice examples of groups which are not obviously groups
Another example of a subgroup that is not obviously a subgroup is given by Frobenius conjecture, for which the only known proof for relies on the classification of finite simple groups. According to Frobenius conjecture, if $n$ divides the order of $G$ and there are exactly $n$ solutions to $x^n = 1$, then the set of solutions forms a subgroup. Let $G$, $H$ and $N$ be as in your answer. You can prove that $N = \{x \in G: x^{[G:H]} = 1\}$ and that $N$ contains exactly $[G:H]$ elements. So your answer can be seen as a particular case of Frobenius conjecture.
Apr
14
answered Which groups are derived subgroups?