Let $H$ be a proper subgroup of maximal order in a group $G$. Now suppose $H$ is solvable and normal in $G$ so there exists a series of normal subgroups:  $1 = H_0 ⊲ H_1 ⊲ ... ⊲ H_n = H$, with $\dfrac{H_{j+1}}{H_j}$ abelian. Consider the cyclic group $&lt;g&gt;$ = $K$ generated by some $g\in G-H$. Now $HK = G$ by maximality of $H$. Thus $\dfrac{G}{H}\cong\dfrac{HK}{H}\cong\dfrac{K}{H∩K}$ wich is a quotient of an abelian group ($K$ is cyclic and hence abelian) so $G/H$ is abelian. We can now construct a series of normal subgroups as following: $1 = H_0 ⊲ H_1 ⊲ ... ⊲ H_n = H ⊲ G = H_{n+1}$  with $\dfrac{H_{j+1}}{H_j}$ abelian. So G is solvable.   My question is, is there another way to approach this proof or in some way generalize it?   Can we in some way have less restrictions on $H$ and still make the proof work for example?
36498
I recently read the well known theorem that for a group $G$ and $H$ a normal subgroup of $G$, then $G$ is solvable if and only if $H$ and $G/H$ are solvable. In my book, only the fact that $G$ is solvable implies $H$ is solvable was proven. I was able to show that if $H$ and $G/H$ are solvable, then so is $G$, but I can't quite show that $G$ is solvable implies $G/H$ is solvable.   My idea was this. Since $G$ is solvable, there exists a normal abelian tower $$ G=G_0\supset G_1\supset\cdots\supset G_r=\{e\}. $$ I let $K_i=G_i/(H\cap G_i)$, in hopes of getting a sequence $$ G/H\supset K_1\supset\cdots\supset K_r=\{e\}. $$ My hunch is that the above is also a normal abelian tower. However, I'm having trouble verifying that $K_{i+1}\unlhd K_i$ and that $K_i/K_{i+1}$ is abelian.  Writing $H_i=H\cap G_i$, I take $gH_{i+1}\in K_{i+1}$ for some $g\in G_{i+1}$. If $g&#39;H_i\in K_i$, then I want to show $g&#39;H_igH_{i+1}g&#39;^{-1}H_i$ is still in $K_{i+1}$, but manipulating the cosets threw me off. I also tried to use either the second or third isomorphism theorems to show that $K_i/K_{i+1}$ is abelian, but I'm not clear on how to apply it exactly. I'd be grateful to see how this result comes through. Thank you.
98497
So I tried looking everywhere but I'm really lost with this proof. I need to to show that:  Let $H$, $K$ be normal subgroups of a group $G$, satisfying $H\cap K = \{e\}$. Let $h\in H$, $k \in K$. (a) Prove that the commutator $w = hkh^{-1}k^{−1}$ satisfies $w \in H$ and $w \in K$. Then conclude that $w = e$ (b) Prove that $hk = kh$, for all $h \in H$, $k \in K$.
3502
I know that given a group $G$ and two normal subgroups $H,K \subset G$ then it is not true that:  "if $H \cong K$ then $ \frac{G}{H} \cong \frac{G}{K} $ (the counterexample is quite easy with products of cyclic groups) "  My question is: Is the converse true?   i.e.     Given that $\frac{G}{H} \cong \frac{G}{K}$ then $H \cong K$ ?   I feel that the answer is no, but I can't think of an example.
92665
This is problem 3B.12. from the afore mentioned book. It asks to show that, if $G$ is a solvable Frattini-free group (that is, $Φ(G)=1$) and $H$ is a subgroup of some maximal subgroup $M$ of $G$, then $G$ contains a subgroup of index $|M:H|$. It is easy to see that, if   $N \unlhd G$ and $M\cap N=1$, then $G=MN$, $M\cong G/N $ and $H\cong K/N $ for some $ K \leqslant G $, where $K$ contains $N$. Then $|G:K|=|G/N:K/N|=|M:H| $ and we are done. However, the hypothesis will not necessarily hold. But, if $N$ is minimal among normal subgroups of $G$ (in the sense of containment), then $N$ is elementary abelian, hence nilpotent, which implies that $M \cap N \lneq N_G(  M \cap N ) \cap N $. Thus, $M \lneq N_G(  M \cap N )$ and by the maximality of $M$ we deduce that $M \cap N \unlhd G$. So, for this particular $M$, if it is not true that there exists some normal subgroup of $G$ with which it intersects trivially, then $M \cap N = N$ by the minimality of $N$. Hence $M$ contains every minimal subgroup of $G$.     How should I proceed from here?
22984
Let $G$ be a finite abelian $p$-group. It is quite elementary to see that if $g \in G$ is an element of maximal order (and thus its span is a cyclic subgroup of $G$ of maximal order) then $G$ can be written as the direct sum $G=\langle g \rangle \oplus H$ for some $H \leq G$ (subgroup of $G$). For a proof see  for example (page 2).  My question: Do we need that $G$ is a $p$-group or does it also work for arbitrary finite abelian groups?   I think it is wrong for general groups because I looked around quite a bit and always only found the above theorem, but I could not find a counter-example.
102548
Let $G$ be a finite abelian group and let $\phi: G \rightarrow G$ be a group homomorphism. I am trying to show that there is a positive integer $n$ such that $G \cong \ker(\phi^{n}) \times \phi^{n}(G)$.  I know that since $G$ is abelian we have that $\ker(\phi^{k})$ and $\phi^{k}(G)$ are normal subgroups of $G$ for any $k$  I also suspect we have the following towers which stabilize at $m$ and $m'$  $$\phi(G) \unrhd \phi^{2}(G) \unrhd \cdots \unrhd \phi^{m}(G)=\phi^{m+1}(G)=\cdots$$  $$\ker(\phi) \unlhd \ker^{2}(\phi) \unlhd \cdots \unlhd \ker^{m'}(\phi)=\ker^{m'+1}(\phi)=\cdots$$  I know that if given a group of the form $HK$ where $H$ and $K$ are normal subgroups of $HK$ and $H\cap K=1$ then $H \times K \cong HK$. I want to apply this to this situation but I am unable to show how write $G$ as a product $HK$  Resolution One of the comments directed me to the following article:  The last part of which answers the question. Below is what is says:  Choose $n=\max(m,m')$, then we have for $x \in \ker^{n}(\phi) \cap \phi^{n}(x)$, this means that $x=\phi^{n}(y)$ for some $y \in G$. This gives:  $0=\phi^{n}(x)=\phi^{2n}(y)$ which means that $y \in \ker^{2n}\phi=\ker^{n}{\phi}$. and then we have that $0=x=\phi^{n}(y)$.  The answer below kindly points out that every element $x$ is contained in one of the cosets of $G/\ker^{n}(\phi)$, this means that $x=k+g$ for some $k \in \ker^{n}(\phi)$ and $g \in G$. We also have that $g=\phi^{n}(h)$ for some $h \in G$. Writing $x=k+\phi^{n}(h)$ essentially shows that $G=\ker(\phi^{n}) + \phi^{n}(G)$. Using the fact above, the claim follows.
80770
Just wanted some feedback to ensure I did not make any mistakes with this proof. Thanks!  Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$, and $|G| &gt; 1$, so for some $x\in G$ we have $x\neq 1$ and $\langle x\rangle\leq G$, so $\langle x\rangle = G$. Suppose $x$ has either infinite order or even order. Then $\langle x^2\rangle \leq G$ but $\langle x^2\rangle \neq \langle x \rangle$, a contradiction. So $x$, and therefore $G$, has odd order, and by Feit-Thompson, $G\cong Z_p$ for some prime $p$.  Edit: Thanks, I see that Feit-Thompson is too much.  Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$, and $|G| &gt; 1$, so for some $x\in G$ we have $x\neq 1$ and $\langle x\rangle\leq G$, so $\langle x\rangle = G$. Suppose $x$ has infinite order. Then $\langle x^2\rangle \leq G$ but $\langle x^2\rangle \neq \langle x \rangle$, a contradiction. So $x$, and therefore $G$, has finite order. Suppose $x$ has composite order $n$ so for some $p &gt; 1$ that divides $n$, $\langle x^p \rangle$ is a proper non-trivial subgroup of $G$, so $G$ is not simple. So $G$ is a cyclic group of prime order.
79950
$G$ is a Group of order $pq$, if $G$ has exactly one subgroup of order $p$ and another with order $q$, then $G$ is cyclic ($p,q$ are prime)  Here's what I have so far, I'm not sure if it is correct.  Let $H, K$ be subgroups of $G$ $|H| = p$  $|K| = q$   We know that by Lagrange's Theorem:  $\forall a \in H : |a|=p$ where $a \neq e$ $\Rightarrow H = &lt;a&gt;$  Similarly: $\forall a \in K : K = &lt;a&gt;$  We also know that: $\forall a \in G : |a| | pq \Rightarrow |a| = p$ or $|a| = q$   $|a| = p \Rightarrow a \in H$ since there is exactly one subgroup of order $p$  $|a| = q \Rightarrow a \in K$ for similar reasons  This means every element in $G$ is either in $H$ or $K$, which are both cyclic. That means $G$ is cyclic.  What am I missing here? Have I made any mistakes and/or is the proof incomplete?
118880
Problem: Let $G$ be a group of order $108 = 2^23^3$. Prove that $G$ has a proper normal subgroup of order $n \geq 6$.  My attempt: From the Sylow theorems, if $n_3$ and $n_2$ denote the number of subgroups of order $27$ and $4$, respectively, in $G$, then $n_3 = 1$ or $4$, since $n_3 \equiv 1$ (mod $3$) and $n_3~|~2^2$, and $n_2 = 1, 3, 9$ or $27$, because $n_2~|~3^3$.  Now, I don't know what else to do. I tried assuming $n_3 = 4$ and seeing if this leads to a contradiction, but I'm not even sure that this can't happen. I'm allowed to use only the basic results of group theory (the Sylow theorems being the most sophisticated tools).  Any ideas are welcome; thanks!
67287
Let $G$ be a group and $H$ be a subgroup of $G$ with finite index. I want to show that there exists a normal subgroup $N$ of $G$ with finite index and $N \subset H$. The hint for this exercise is to find a homomorphism $G \to S_n$ for $n := [G:H]$ with kernel contained in $H$.   The standard solution suggests to choose $\varphi$ as the homomorphism induced by left-multiplication $\varphi: G \to S(G/H) \cong S_n$. I'm not 100% sure if I understand this correctly. What exactly does $\varphi$ do? We take $g \in G$ and send it to a bijection $\varphi_g: G/H \to G/H, xH \mapsto gxH$? If so, how can I see that its kernel is contained in $H$? Also, the standard solution claims its image is isomorphic to $G/N$ and thus $N$ has a finite index in $G$, how can I see that the image is isomorphic to $G/N$?  Thanks in advance for any help.
40822
Let $G$ be a non-nilpotent and supersoluble finite group.  $G=Q\ltimes P$ and $|Q: C_{Q}(P)|=q$ is prime, Where $P$ is a group of prime order $p\neq q$.   $Q$ is noncyclic group of order $|Q|=q^{\alpha}$ where $\alpha &gt;2$ and all maximal subgroup of $Q$ diffrent from $C_{Q}(P)$ are cyclic Then every $2$-maximal subgroup of $G$ permutes with all $3$-maximal subgroup of $G$.  I dont have idea ?  What i do? Maximal subgroup of $G$ is not clear for me.    A subgroup $H$ of a group $G$ is called $2$-maximal (or second maximal )in $G$ if $H$ is a maximal subgroup of some maximal subgroup $M$ of $G$.
36159
Is there a better (more direct or intuitive) proof for this proposition than I have come up with below? I am not sure whether it could be simiplified:  Let $G$ be a group with $H \leq G$. Then $K = \bigcap_{g \in G} gHg^{-1}$ is normal in $G$.  Let $a \in K$. Then $a \in gHg^{-1}$ for all $g \in G$. Therefore for all $g_1,g \in G$, $g_1ag_1^{-1} \in g_1gHg^{-1}g_1^{-1} = (g_1g)H(g_1g)^{-1}$ and so $g_1ag_1^{-1} \in K$ since $g_1g \in G$. Then $K$ is normal in $G$.
51723
Suppose $H$ is the only subgroup of order $o(H)$ in the finite group $G$. Prove that $H$ is a normal subgroup of $G$.  I've been trying this problem for quite a while but to no avail. What I can't understand is, how do you relate the subgroup being normal/abnormal to its order?  This question is from I.N.Herstein's book Topics in Algebra Page 53, Problem no. 9. This is NOT a homework problem!! I'm studying this book on my own.
43458
I think I remember seeing a theorem about this in my Abstract Algebra class, but I cannot seem to find the reference for it anymore.   Let $G$ be a group and let $H$ and $K$ be non trivial subgroups of $G$. Are there any sufficient conditions on $H, K$ and $G$ such that $G = HK$? (In other words $\forall g \in G$, there exist $h \in H$ and $k \in K$ such that $g = hk$)
60423
If $H$ be a subgroup of a group $G$  such that $Ha \not=Hb$ implies that $aH\not=bH$.Then how can I show that $gHg^{-1}\subset H$  $\forall$ $g\in G$?  I do not think I have made any progress.However, this is what I have done: For any $h\in H$,$(ghg^{-1})(gh^{-1}g^{-1})=e$ where $e$ is the identity element of the group $G$.  And for $h,k\in H$, $(ghg^{-1})(gkg^{-1})=g(hk)g^{-1}$ which is in $gHg^{-1}$ as $hk\in H$. Thus $gHg^{-1}$ is a group.  I do not think I am getting anywhere.
109600
I was having a discussion with a friend of mine about some normal group properties and then came up with the question "if G is not commutative, then is there always a subgroup that is not a normal subgroup?" It's probably more easy to solve this in the following form:$$\forall H \leq G : H  \lhd G \Rightarrow \forall a,b \in G : ab=ba$$ My question is, can anybody give a proof, or a counter-example (because I don't think it holds) of this theorem? Thanks!
70921
Prove that if H has finite index n then there is a normal subgroup N of G with $N \subset H $ and [G:N]$ \le $n!.  I tried to solve the problem but could not done exactly. Since [G:H]=n , Let A={$a_i H$:i=1,2,..,n}. Consider the left multiplication group action on A .Then kernal of the action is intersection N= $\cap a_iHa_i^{-1}$ i=1,2..,n.  Then N is the largest normal subgroup of G contained in  H. Now it is clear that $[G:a_iHa_i^{-1}]$ =n for all i.{by orbit stabliser theorem}.Now how to find [G:N]? yes I got it. By Caley's theorem G/N is isomorphic to a subgroup of $ S_{A}$ .Therefone [G:N] divides n!.