I have one problem, that I think it's pretty hard to solve. The problem reads:     Let $G$ be a finite $p-$group, and $H &lt; G$ (i.e $H \le G$, and $H \neq G$). Prove that $H \neq N_G(H)$.   Here are my thoughts on this problem:     I know that if $G$ is a finite $p-$group, then its center is non-trivial, i.e $\mathcal{Z}(G) \neq \{e\}$. Then, I start off by Proof by Contradiction (note that we already have $H \lhd N_G(H)$), i.e, assume that $H = N_G(H)$, then for every $g \notin H$, there exists $h_g\in H$, such that $g^{-1}h_gg \notin H$. And from here, I come up with the fact that, in order for $H$ to satisfy $H = N_G(H)$, we must have $\mathcal{Z}(G) \lhd H$.   However, the two facts above don't seem to help much. :(  Can you guys give me a little push on this problem?  Thanks so much,  And have a good day,
123852
If $o(G) = p^n,$ p a prime number, and H is a subgroup of G, show that there exists an $x\in G$, but $x\notin H$ such that $x^{-1}Hx = H$  How can I prove this?
2653
Problem Suppose G is a finite group of order $n$ which has a unique subgroup of order $d$ for each $d\mid n$. Prove that $G$ must be a cyclic group.  My idea:   I try to prove it by induction. Let $p|n$ be a prime. Then by condition, there exists a unique subgroup $H$ of order $n/p$. Since $|gHg^{-1}|=|H|$, we must have $gHg^{-1}=H$ by the uniqueness part of the condition. So, H is a normal subgroup. Now, $|G/H|=p$ and thus $G/H=\langle x\rangle$ where $x^p \in H$.    However, I cannot continue.  My another idea is that first consider the case when $|G|$ is a power of some prime $p$. But, it still doesn't work.
67137
I am to trying to prove this theorem: A finite $p$-group cannot be simple unless it has order $p$.  I have this:   Let $G = P$ and $|G|=p$; then there exists $N$, a normal subgroup of $G$ by Lagrange's theorem, such that $|N| \mid |G|$ so, $|N|= |e| = 1$ or $|N| =  p$  If $|N|=p$ so $N=G$.  But I don't know how I can prove the part in the case of $|N|=|e| = 1$  Is my start correct?
30101
Suppose $G=H\cup K$, where $H$ and $K$ are subgroups. Show that either $H=G$ or $K=G$.  What I did: For finite $G$, if $H\neq G$ and $K\neq G$, then $|H|,|K|\le |G|/2$. But they clearly share the identity element, so $H\cup K\neq G$.  How can I extend it to $G$ infinite?
79390
I have this written down in my notes, but cannot seem to find verification of it. How can I see this fact?   I believe there is a theorem, something along the lines of if a group G is isomorphic to H then, H is a group. Can anybody verify this or point me to the direction I am looking for? Thanks.
43912
Assume that $G$ is finite with $p$ the smallest prime dividing its order. Suppose $H &lt; G$ with $[G:H]=p$. Prove that $H \lhd G$.  I've seen this question a few times on here but all the proofs I saw appeal to the First Isomorphism Theorem. I had old class notes showing this proved solely with group actions and orbits but I could only make out some of it so I recreated it from scratch. Can you guys check this for me?  Let $H$ act on the set $G/H$ by left multiplication. Consider the orbit of $H$ itself. $\mathcal{O}_H=\{hH \mid h\in H\}$ and $hH = H$, thus $\mathcal{O}_H$ contains one coset, namely, $H$. Let $x\in G - H$. What can be said about $\mathcal{O}_{xH}$? This action of left multiplication partitions $G/H$ into disjoint orbits of cosets. Because $|G/H| = p$, the sum of the sizes of the orbits is equal to $p$. We already know that $|\mathcal{O}_H| = 1$ therefore       $|\mathcal{O}_{xH}| \leq p-1$. The Orbit-Stabilizer theorem gives us  $|\mathcal{O}_{xH}| |Stab_{xH}| = |H|$. We see that $|\mathcal{O}_{xH}|$ divides $|H|$ and hence divides $|G|$. But since $p$ was the smallest prime dividing the order of $G$, $|\mathcal{O}_{xH}| = 1$.  The immediate consequence is that $|Stab_{xH}| = |H|$. As $Stab_{xH} \leq H$, we have $Stab_{xH} = H$. Therefore for every $h \in H$, $hxH=xH$. This implies  $x^{-1}hx\in H$ and thus $x^{-1}Hx = H$ for all $x \in G$. Therefore, $H \lhd G$.
108605
Let $G$ and $G'$ be groups. Let $\phi: G \to G'$ be a homomorphism. Let $H$ be a group of $G$ and let $H'=\{\phi(h): h\in H\}$.  Want to show that $H'$ is a subgroup.  Proof:  (i)  $H'$ has identity since $\phi(e_G)\in H'$.  This is because $\phi(e_G)=e_{G'}$.  (ii)  $H'$ is closed because of the following: Let $h_{1}^{'}, h_{2}^{'} \in H'$.  Then there exists an $h_1,h_2$ in $H$ such that $h_1^{'}=\phi(h_1)$ and $h_2=\phi(h_{2}^{'})$.  Then $h_1^{'}h_2^{'}=\phi(h_1)\phi(h_2)=\phi(h_1h_2)$.  (iii) For each $h^{'} \in H'$ there exists an $h \in H$ such that $\phi(h)=h'$.  But, $h'^{-1}=\phi(h)^{-1}=\phi(h^{-1})$.  Hence it has inverses.  Did I prove this correctly?
3395
I'm having trouble with an exercise in "Introduce to the theory of group" book. This is problem:  Let $M$ be a maximal subgroup of $G$. Prove that if $M$ is a normal subgroup of $G$, then $[G: M]$ is finite and equal to a prime.
101715
Suppose that $ N $ is a normal subgroup of a finite group $ G $. If $ |N| = 5 $ and $ |G| $ is odd, why is $ N $ contained in $ Z(G) $, the center of $ G $?  I know how to do this when $ |N| = 2 $ and $ |G| $ is even, but am not sure what to do with this one. Sylow's theorems maybe?  Thank you guys!
61950
So if I'm given $H$, when I can conclude there is a group $G$ such that $H\cong G/Z(G)$? It's easy to show that non-trivial cyclic groups are not of this form. More generally, any group with the property that all finitely generated subgroups are cyclic (e.g. $\mathbb{Q},\mathbb{Q}/\mathbb{Z}$) cannot be of this form. Are there are any others?
57017
Let $G$ be a finite group and $H$ a subgroup of index $p$, where $p$ is a prime. If $\operatorname{gcd}(|H|, p-1)=1$, then $H$ must be normal. Does somebody have a quick proof of this?
37197
Let $H$ be a subgroup of a group $G$ such that $x^2 \in H$, $\forall x\in G$. Prove that $H$ is a normal subgroup of $G$.    I have tried to using the definition but failed. Can someone help me please.
100999
If $H \leq Z(G) \leq G$, where G is a finite group, $Z(G)$ is its center, and $(G:H) = p$ for some prime $p$, then G is abelian.  Well because H &lt; Z(G) we have that H commutes with all of G. Also |G/H|=p. I was thinking of connecting this result to the fact that if $G/Z(G)$ is cyclic, then $G$ is abelian, but was unable to. Then I thought about using the result that $(G/H)/(Z(G)/H) = G/Z(G),$ but that leads to nowhere. Any suggestions?
33615
Let $G$ be a finite group and $N$ be a normal subgroup of $G$ such that $G/N$ is nilpotent.  a) Prove that there exists a nilpotent subgroup $K$ of $G$  such that $G=NK.$  b) Suppose that $N$ is abelian and $Z\left(G\right) =\lbrace e \rbrace.$ Show that if $K$ is nilpotent and $G=NK$ then $K=N_G\left(K\right)$ and $K \cap N =\lbrace e \rbrace.$  Could anyone help me to give a solution for the problem! thank you!
66543
I need some help in relation to this exercise    "Prove that if a normal subgroup $H$ of $ G$ has index $n$, then $g^n \in H$ for all $g \in G$."  I'm not allowed to use quotient groups in the proof, because the exercise is in the chapter before.   I tried by induction on $n$. The case $n=1,n=2$ are obvious, but even  the case $n=3$ is giving me trouble so I give up studying the general case of the inductive step.  My other approach was studying left or right coset of $G$. But I only proved that if $g \in aH$ then $g^2 \notin aH$ if $a \notin H$, and I can't find a way to demonstrate that $g^n \in H$. (my starting idea was to prove that every power of $g$ is in a different coset but then I realize that in this way I don't handle several case, for example $g$ has period strictly lesser than $n$ and in conclusion it doesn't prove the exercise)  Maybe I'm missing something about indexes, and this is why I asked here for some help,  (I can't use quotient groups because they are introduced later than this exercise, forgot to add this info in the beginning) Thanks in advance :)
18939
Let $H \leq G$. If $G = H H^x$ then $G=H$.  According to the source, it should solve with elementary knowledge of group theory, but my tries have no result! Any help would be appreciated.   Thanks in advance.
6550
If $G$ is a finite $p$-group with a nontrivial normal subgroup $H$, then the intersection of $H$ and the center of $G$ is not trivial.