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2d
comment A problem on order of a Group.
In general. Not only for finite groups. Proof: (sketch) assume $H \leq G$ has two (right) cosets, then we have a disjoint union $G=H \cup Hg$, $g \notin H$. Let $h \in H$. If $g^{-1}hg=kg$ for some $k \in H$, then $g=hk^{-1} \in H$, a contradiction. So $g$ normalizes $H$, and $H$ is normal.
2d
comment A problem on order of a Group.
Yes every subgroup of index $2$ is normal. One of the first things you learn on Group Theory.
2d
revised A problem on order of a Group.
added 16 characters in body; edited title
2d
answered A problem on order of a Group.
Jun
25
revised Let $G$ be a group containing exactly $2n$ elements, $n\ge1$ integer.
edited tags
Jun
24
comment Possible Class equation for a group
A nice proof of Bjorn Poonen has been published in the Am. Math. Monthly 105 (1998) nr 4.
Jun
23
answered On conjugacy class size of an element and its order.
Jun
23
answered If G is a group of order n=35, then it is cyclic
Jun
19
comment Origin of the term “derived subgroup”
There is also a connection to Lie algebra's and brackets.
Jun
18
revised Definition of finite-by-nilpotent
added 4 characters in body
Jun
17
answered Definition of finite-by-nilpotent
Jun
14
revised If $C_G(H)=N_G(H)$ for all abelian subgroups, prove that $G$ is abelian
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Jun
14
comment If a normal subgroup $N$ of order $p$($p$ prime) is contained in a group $G$ of order $p^n$,then $N$ is in the center of $G$.
You are welcome!
Jun
14
revised If $C_G(H)=N_G(H)$ for all abelian subgroups, prove that $G$ is abelian
added 2 characters in body
Jun
14
revised If $C_G(H)=N_G(H)$ for all abelian subgroups, prove that $G$ is abelian
edited title
Jun
14
answered If $C_G(H)=N_G(H)$ for all abelian subgroups, prove that $G$ is abelian
Jun
14
revised If $C_G(H)=N_G(H)$ for all abelian subgroups, prove that $G$ is abelian
added 27 characters in body
Jun
13
comment find the number of elements of order 3 in an abelian group of order 120
That is a coincidence. For example, if the Sylow subgroup would be isomorphic to $C_9$, the number of elements of order $3$ is $2$ while $\varphi(9)=6$, being the number of elements of order $9$.
Jun
13
comment find the number of elements of order 3 in an abelian group of order 120
Each Sylow $3$-subgroup here has order $3$. This implies that each pair of distinct Sylow $3$-subgroups intersect trivially. Each of them contains two elements of order $3$. So this explains the total number of elements of order $3$ being $2n_3$. In general a Sylow subgroup is more than just a cyclic group of prime order and there is no nice formula for the total number of elements of a particular order. However, the number of elements of order $p$ is always $\equiv -1$ mod $p$.
Jun
12
answered If a normal subgroup $N$ of order $p$($p$ prime) is contained in a group $G$ of order $p^n$,then $N$ is in the center of $G$.