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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
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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?
Apr
13
comment What are central automorphisms used for?
Some basic facts. You can prove that $\theta \in \operatorname{Aut}(G)$ is a central automorphism if and only if $\theta$ commutes with every inner automorphism. Thus $\operatorname{Aut}_c(G)$ is the centralizer of $\operatorname{Inn}(G)$ in $\operatorname{Aut}(G)$. Also, there is a bijection between $\operatorname{Aut}_c(G)$ and $\operatorname{Hom}(G/G', Z(G))$.
Apr
10
comment On order automorphisms group of a finite group
The smallest examples of groups with nontrivial automorphism groups of odd order are of order $3^6$. For example, $G = \operatorname{SmallGroup}(3^6, 90)$ has an automorphism group of order $3^7$. So by your remark, $C_2 \times G$ is a smallest possible example of even order. Reference: D. MacHale and R. Sheehy, Finite groups with odd order automorphism groups, Math. Proc. R. Ir. Acad. 95A (1995), no. 2, 113–116. JSTOR
Apr
9
revised Finite group with a normal Sylow subgroup
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Apr
9
answered Finite group with a normal Sylow subgroup
Apr
6
comment A group of order $195$ has an element of order $5$ in its center
@wqr: $G'$ is the commutator subgroup of $G$, often also denoted by $[G, G]$. See wikipedia
Apr
6
answered Is there an idempotent element in a finite semigroup?
Apr
5
revised Proving that a group of order $112$ is not simple
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Apr
5
revised Proving that a group of order $112$ is not simple
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Apr
5
comment Proving that a group of order $112$ is not simple
It's up to you. If you keep it maybe you should edit in a note about the mistake so it's not hidden in the comments. I'd say keep it, I think this is a common pitfall when doing these sorts of counting arguments related to Sylow subgroups.