Show that if $G$ is a group with $|G| = p^nq$, with $p$ and $q$ are primes and $p &gt; q$, then $G$ contains a unique normal subgroup of index $q$.
122260
I'm trying to prove if $|G|=p^nq$ with $p\gt q$, primes, then $G$ contains a unique normal subgroup of index $q$.  I know by the first Sylow theorem that G has a Sylow p-subgroup $P$ with $[G:P]=q$. My problem is prove that it's unique.  I need help  Thanks in advance
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.
110998
I'm trying to understand the proof that any group of index $2$ is normal.   Let $G$ be a group and $H$ a subgroup of index $2$.  I understand that if $x\in H$ then $xH=H=Hx$ since $H$ is a subgroup. However, I don't understand why if $x\in G\setminus H$ then $xH=G\setminus H.$  I know there are other proofs of this on this site but none of them specifically address this question.
106325
Let $G$ be a group such that  $G'$ abelian  and any abelian normal subgroup of $G$ is  finite. Show that $G$ is finite.
86983
Let $K$ be a normal subgroup of $G$ of index $m$. If $n$ is a positive integer such that $\gcd(m,n)=1$, show that all the elements of order $n$ are in $K$.   I could solve this if $G$ is finite as follows: Let $g$ be an element of order $n$. $|gK|$ divides $|G|$ using canonical epimorphism, again $|gK|$ divides $|G/K|$. Since $m$ and $n$ are relatively prime, we get order of $gK$ is $1$.  What if $G$ is infinite?
92045
Let $G$ be a finite group and $N$ a normal subgroup of $G$ of index $4$. Prove that $G$ has a subgroup of index $2$ (hence normal) and if $G/H$ is not cyclic then $G$ is equal to the union of three proper subgroups.   I understand that $G/H$ is isomorphic to $\mathbb{Z}_4$ or $\mathbb{Z}_2\times \mathbb{Z}_2$. Let $K^*$ be a subgroup of $G/H$ of order $2$. Then $K^*\cong K/H$ for some $K$ such that $H$ is normal in $K$ and hence $[G:K]=[G/H:K/H]=4/2=2$.  If $G/H$ is not cyclic then $G/H\cong \mathbb{Z}_2\times \mathbb{Z}_2$. I don't know how to prove $G$ is the union of three subgroups. Could it be true that as $G/H$ is the union of three proper subgroups $K_1^*,K_2^*,K_3^*$ then the corresponding groups $K_1,K_2,K_3$ have union equal to $G$? Would anyone give me a hint to prove this?  Thank you.
22464
I would be very thankful if someone could give me a hint with proving this. It's a very common exercise in abstract algebra textbooks.     If $G$ is a group with a subgroup $H$ of finite index $n$, then $G$ has a normal subgroup $K$ contained in $H$ whose index in $G$ is finite and divides $n!$.   I found the proof on  at some point (although the proof appears to be no longer there), but I got lost in one of the details.
122933
Suppose $G$ is a group and has exactly two nontrivial proper subgroups. Prove that $G$ is cyclic and $|G|=pq$ where p,q are distinct primes or $G$ is cyclic and $|G|=p^3$ where $p$ is a prime. Usually my tactic to show a group is cyclic is find an elements such that its order and order of the group are the same. But here I cannot do so because I have no informatio regarding order of elements. So I stuck. Can anyone help me?
2057
Respected all  Please help me on this problem.   Suppose that $G$ is given group and $N$ be a normal subgroup of $G$ and $N$ is isomorphic to $G_1$ as groups. If $G/N$ is isomorphic to the group $G_2$, can we establish that $G$ is isomorphic to the group $G_1\times G_2$ ?
87453
These questions are really bothering me. Any help would be much appreciated.  Let $p$ be prime  a) If $(G, \oplus)$ has order $2p$, prove that every proper subgroup of $(G, \oplus)$ is cyclic  b) If $(G, \oplus)$ has order $p^2$, prove that $(G, \oplus)$ has a subgroup of order $p$
63316
Let $\phi:G\to G'$ be a group homomorphism, and let $N'$ be a normal subgroup of $G'$. Show that $\phi^{-1}[N']$ is a normal subset of $G'$.  My attempt: $\phi^{-1}[N']=\{g\in G:\phi(n)\in N'\}$ $$g' n' (g')^{-1}\in N'$$ $$\Rightarrow\phi^{-1}(g' n' (g')^{-1})=\phi^{-1}(g')\phi^{-1}(n')\phi^{-1}(g'^{-1})$$ Thus $\phi^{-1}[N']$ is normal. Is this thinking correct? I feel like I am missing something.
112790
Im having trouble with the second part of this question,  Let $H$ be a normal subgroup of $G$ with $|G:H| = n$,  i) Prove $g^n \in H$ $\forall g \in G$ (which i have done)  ii) Give an example to show that this all false when $H$ is not normal in $G$.(which I am having trouble with showing)  Any suggestions?
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!.
42433
I'm trying to figure out this problem:  Problem: Let $G$ be a group of order $p^k q$ with $p$ and $q$ distinct primes and $k \in \mathbb{N} \setminus \left\{0\right\}$.   a) Prove: $G$ has a subgroup of order $p^k$, or $p$ divides $|Z(G)|$. (Here $Z(G)$ means the center of $G$).   b) Prove: if $p$ divides $| Z(G)|$, there there exists an $x \in Z(G)$ of order $p$.   Attempt: For a) I tried proceeding by induction.  Base step: Assume $k=1$. Then $|G| = pq$ with $p \neq q$. Case 1: if $G$ is commutative, then $Z(G) = G$ and then $p$ divides $|Z(G)|$.   Case 2: Assume $G$ is not commutative. Then the center of $G$ is trivial, i.e. $Z(G) = \left\{e_G\right\}$ (this is a theorem of group theory). So $p$ does not divide $|Z(G)|$. Then we must show that $G$ has a subgroup of order $p$. Let $H$ be a non-trivial subgroup of $G$, that is $H \neq G$ and $H \neq \left\{e\right\}$. Then by Lagrange, $|H|$ divides $pq$. So we must have either that $|H| = p$ or $|H| = q$ (is this correct? Since $H$ is non-trivial). If $|H| = p$ then we are done. If $| H | = q$, then $|G / H| = p$ and so $G/H$ is cyclic. I want to derive a contradiction from this, and so show that $H$ cannot have order $q$, but I don't know how.  Induction step: Assume the claim holds for $|G| = p^{k-1} q$. We must then show that it also holds for $|G| = p^k q$.  I want to use the class equation. We have that $$ p^k q = |Z(G)| + \sum_{i=m+1}^n |Cl(a_i)| $$ where $|Cl(a_1)| = \ldots = |Cl(a_m)| = 1$ and $|Cl(a_i)| &gt; 1$ for $i &gt; m $. Since $p$ is a divisor of the LHS, it must also divide the linear combination. But how can I conclude from this that $p$ divides $| Z(G)|$ ? I don't know how to use the induction hypothesis.   b) Assume $p$ divides $|Z(G)|$. I first assume that $G$ is abelian. Then by structure theorem for finite abelian groups, we know that $$ G \cong \mathbb{Z}_{p_1^{k_1}} \oplus \ldots \oplus \mathbb{Z}_{p_r^{k_r}}. $$ This means that $$ p^k q = p_1^{k_1} \ldots p_r^{k_r}. $$ So $p | (p_1^{k_1} \ldots p_r^{k_r})$ and so since $p$ is prime, $p$ must divide at least one factor. Assume that $p$ divides $p_i^{k_i}$. Since $p_i^{k_i} = p_i \ldots p_i$ with $k_i$-amount of factors, we must again have that $p$ divides $p_i$. So $p = p_i$.   Now, since every $\mathbb{Z}_{p_i^{k_i}} = \mathbb{Z}_p^{k_i}$ is abelian, we have that the center equals the whole group. Let $x_i$ be an element in this center. Then $order(x_i) | p^{k_i}$ and so $order(x_i) | p$. Hence the order of $x_i$ equals $p$ in $\mathbb{Z}_p^{k_i}$. Since $x_i$ is also in $Z(G)$, this concludes the proof?   And what about the case that $G$ is not abelian?  Help with this problem is appreciated!
64989
I'm struggling to proof that if $H$ and $K$ are subgroups of  finite index of a group $G$ such that $[G:H]$ and $[G:K]$ are relatively prime, then $G=HK$. I don't know why I can't answer it, because this question seems easy. I'm stuck maybe because I've studied so far just Lagrange's theorem and some of its consequences. But I think we don't need much more, because this is the material covered so far by the Hungerford's book.  I need help.  Thanks.
117977
I am doing the following exercise from [Birkhoff and MacLane, A survey of modern algebra]:     Let $G$ be a group of order $pq$ ($p,q$ primes).  Show that either $G$ is cyclic or contains an element of order $p$ (or $q$).  In the second case, show that $G$ contains either $1$ normal or $q$ conjugate subgroups of order $p$.  In the latter case the $pq-(p-1)q=q$ elements not of order $p$ form a normal subgroup.  Infer that $G$ always has a proper normal subgroup.   I think I was able to prove $G$ has a proper normal subgroup, but without quite establishing that the number of subgroups of order $p$ is 1 or $q$.  Here is my solution:    Suppose $G$ is cyclic and $G=\langle x \rangle$.  Then, $\langle x^p \rangle$ has order $q$ and is normal, so we are done.  So suppose $G$ is not cyclic.  Let $x \in G-1$.  By Lagrange's theorem, the order of $x$ is $p$, say.  Let $H:=\langle x \rangle$.  $G$ acts on all subgroups of order $p$ by conjugation.  In this action, the stabilizer of $H$ is the normalizer $N_G(H)$.  Since $N_G(H) \supseteq H$, $|N_G(H)|$ equals $p$ or $qp$.  Hence the orbit containing $H$ has length $q$ or 1 (by the orbit-stabilizer lemma).  If this value is 1, then $H$ is normal, so we are done.  If this value is $q$, then there are $pq-(p-1)q=q$ remaining nonidentity elements in $G$ that are not in any conjugate of $H$.    Case 1: If any of these $q$ elements, say some $y \in G$, has order $p$, then $K:=\langle y \rangle$ is either normal (so we are done) or has $q$ conjugate subgroups (in which case the conjugates of $H$ and of $K$ together contain $q(p-1)+q(p-1)+1&gt;qp$ elements, a contradiction).    Case 2:  If none of the $q$ elements has order $p$, then they all have order 1 or $q$, and hence form a cyclic subgroup $K$ of order $q$.  $K$ must be normal, since otherwise $g^{-1}Kg \ne K$ for some $g \in G$, whereas $K$ already exhausted all the elements of $G$ of order $q$.   Thus, $G$ has a normal subgroup of order $p$ or $q$.  While I have shown the group is not simple, I think I haven't shown yet that the number of subgroups of order $p$ is 1 or $q$.  The proof only establishes that the number of distinct conjugates of $H$ is 1 or $q$.  Any suggestions on how to prove (using elementary methods) that any other subgroup of order $p$ would have to be conjugate to $H$? In other words, I need to rule out Case 1 in the proof.
58260
Let $H$ and $K$ be normal subgroups of a group $G$ such that they intersect trivially.  Why is it then the case that   $$hk=kh;\;\;\;\; \forall h\in H,\;\forall k\in K?$$