Suppose G is an finite group. Let N be a normal subgroup of G and A be an arbitraty subgroup. Prove that $$\mid AN\mid=\frac{\mid A\mid\mid N\mid}{\mid A\cap N\mid}$$  My initial approach is that I should somehow invoke one of the isomorphism theorems that involves quotient group and then apply another theorem to grasp about its size. Yet I've no idea how to actually approach it, especially with the given condition that $N$ is normal.
17125
Let $H$, $K$ be subgroups of $G$. Prove that $o(HK) = \frac{o(H)o(K)}{o(H \cap K)}$.  I need this theorem to prove something.
35424
I have to prove that if G is a group, Aut(G) is a subgroup of Sym(G), and later that the inner automorphisms group is a normal subgroup of Aut(G). I have time until tomorrow in the morning, but I have no idea what to do, please please please help me.
95965
Let $G$ pe a $p$ group. I have to show that         the number of nonnormal subgroups is divisible by $p$       the number of subgroups differs from the number of normal subgroups by a power of $p$.   Are there any theorem that can help me to prove this ? We have discussed the Sylow theorems but I don't know how to apply them - if those are the theorems I need. (Does this theorem also hold for infinite groups $G$ ?)
85234
How can I prove that, if we have a group G, then subgroup of G generated by all n-th powers of elements from G is normal subgroup of G?
94334
This is question 2.9 #9 from Topics in Algebra by Herstein:     If $o(G)$ is $pq$ where $p$ and $q$ are distinct prime numbers and if $G$ has a normal subgroup of order $p$ and a normal subgroup of order $q$, prove that $G$ is cyclic.   Let $N$ denote the normal subgroup of order $p$ and $M$ denote the normal subgroup of order $q$. Here are a couple things I noted while exploring the problem:   $N$ and $M$ are cyclic, since they have prime order. $G = NM$ [1].   Any hints where to look next?    [1] Since $N,M$ are normal, $NM \le G$. Then $o(NM) \mid o(G)$, so $o(NM)$ is either $1$, $p$, $q$, or $pq$. The first three don't work because they force one of $N$ or $M$ to be $(e)$, so we have $o(NM) = pq$. Therefore, $NM = G$.
32431
Let $G$ be a group and $H$ be a normal subgroup of index $p$ ( a prime ) ; suppose $K$ is a subgroup of $G$ not contained in $H$ , then is it true that $G=HK$ ? ( I know that the fact is true if $p=2$ , but I don't know for the general case . Please help . Thanks in advance )
44531
If $G = \begin{pmatrix} a &amp; b \\ 0 &amp; 1 \end{pmatrix} a,b\in (\mathbb{R}) : a \neq 0$)   and assume G is a group under matrix multpication  Prove that H = ($\begin{pmatrix} 1 &amp; t \\ 0 &amp; 1 \end{pmatrix}: t\in\mathbb{R}$) is a normal subgroup of G  I'm meant to show that H is a normal subgroup by showing that the left coset is equal to the right coset.  So far I have shown the following:  take $g=\begin{pmatrix} a &amp; b \\ 0 &amp; 1 \end{pmatrix}$   then  $gH=\begin {pmatrix}a &amp;at+b \\ 0 &amp;1 \end{pmatrix}$  and take $g'=\begin{pmatrix} a' &amp; b' \\ 0 &amp; 1 \end{pmatrix}$  then $Hg'=\begin{pmatrix} a' &amp; b'+t \\ 0 &amp; 1 \end{pmatrix}$  I don't quite understand how $gH=Hg'$
93910
Let $(G,•)$ be a group in which $(a•b)^3 = a^3•b^3$. Prove that $H=\{x^3 : x \in G\}$ is a normal subgroup of $G$.  I assumed a homomorphism from $G$ to $G$ by $f(x) = x^3$. Now $\operatorname{Im}f$ is $H$. This is where I got stuck. How can I show that here $\operatorname{Im}f$ is a normal subgroup of $G$? Please help.
27938
Let $G$ be a group, and $H$ a subgroup.  $H$'s normalizer is defined: $N(H):=\{g\in G| gHg^{-1}=H \}$.   Prove $N(H)$ is a normal subgroup of G, or give counterexample.  Intuitively it seems to me that this claim is wrong, however, I'm having trouble with finding a counterexample.  Thans in advance for any assistance!
95937
This is not for homework. (I am a grader for a class.)  The case in which $G$ is finite is trivial. (That is, use a corollary to Lagrange's Theorem, and set $[G:H] = \dfrac{|G|}{|H|}$, and similarly for $[H:K]$.) How do you prove this for when $G$ is infinite?   The proof of this statement that I know well tries to show that $[G:K]$ is finite iff $[G:H]$ and $[H:K]$ are finite, but I am not a fan of that proof. For pedagogical purposes, I'm looking for a better proof. (The course uses right cosets by convention.)  The course has just started covering normal subgroups, so things like the Isomorphism Theorems are not known yet.
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?
38784
Let $G$ be a finite group of order $mn$ with $\gcd(m,n) = 1$ and $H$ be a normal subgroup of order $m$. Prove that it is the only subgroup of order $m$. This is somewhat different from this question:  and I don't think we are allowed to use homomorphisms(it is in the next section) I let $K$ be a subgroup of order $m$. If $K = H$, I am done. Otherwise, I supposed $K \neq H$. Based on previous results, I know that $HK = \{hk | h\in H, k \in H\}$ would be another subgroup of $G$ if $H$ is normal. We would have that $\exists k \in K$ such that $k \notin H$ or $\exists h \in H$ such that $h \notin K$. In either case, $|HK| &gt; m$ which may cause problems with some theorems of number theory and Lagrange's theorem but I am unsure of the specifics. I don't know if I can say more about $|HK|$ or whether I am even on the right track. Thoughts?
49386
Suppose $G$ is a group and $K\lhd H\lhd G$ are normal subgroups of $G$. Is $K$ a normal subgroup of $G$, i.e. $K\lhd G$? If not, what extra conditions on $G$ or $H$ make this possible?  Applying the definitions, we know $\{ghg^{-1}\mid h\in H\}=H$ and $\{hkh^{-1}\mid k\in K\}=K$, and want $\{gkg^{-1}\mid k\in K\}=K$. Clearly, the best avenue for a counterexample is if $gkg^{-1}\not\in K$ for some $k\in K$ and $g\in G-H$.  If no such element exists, $\{gkg^{-1}\mid k\in K\}\subseteq K$ implies $\{gkg^{-1}\mid k\in K\}=K$ because if $k'\in K$, $gk'g^{-1}=k\in K\Rightarrow k'=g^{-1}kg\in\{gkg^{-1}\mid k\in K\}$.
115391
Suppose that G is a group and N &lt; G a normal subgroup. Assume that there is no normal subgroup M &lt; G with N &lt; M. Prove that G/N is simple.  This problem is really throwing me for a loop. The concept of quotient groups and how they relate to simple subgroups is confusing me, and so I am not really sure how to begin. If a simple subgroup's only normal subgroup is itself (and the trivial group), then G/N being simple would be proving that when multiplying the entire subgroup N by an element g for all g in G, that the collection of those sets cannot be broken into smaller normal subgroups?
92507
Let $H$ be a subgroup of $G$ with $[G:H]=n$ and $G$ finite.  Let $G$ act on left cosets of $H$ by left multiplication.  Let $N$ be the kernel of the action.  Show that $[G:N]$ divides $n!$.  Ok so I have a theorem that says that if $G$ is simple then it embeds into $S_n$.  So if $G$ is simple than I'm done because $G$ divides $n!$ and thus $N$ divides $n!$ and hence $[G:N]$ divides $n!$.  Now I also know that $N$ is the largest normal subgroup of $G$ contained in $H$.  Thus if $G$ isn't simple then it contains some non-trivial normal subgroup $K$ and thus $K\cap H$ is normal in $H$ and by the 2nd Iso Theorem $H/K\cap H\cong KH/H$...    As you can see I'm not sure where to go from here. I was hoping to prove that $K\cap H$ wasn't trivial and thus that $N$ couldn't be trivial since it must contain $K\cap H$, but I'm not sure I can, and even if I could I'm not sure that would help me in showing that $[G:N]$ is small enough to divide $n!$.  Can anyone give some guidance?  Thanks.
99526
For an arbitrary group $(G,\cdot)$ let $\operatorname{Aut}(G) = \{f: G \to G \mid f \text{ is an isomorphism}\}$ be the set of all automorphisms of the group $G$. We assume that $(\operatorname{Aut}(G),\circ)$ where $\circ$ is composition of mappings is a group.  1) Prove that for arbitrary $a \in G$ there is an automorphism $p_a: G \to G;\quad p_a(x) = a^{-1}xa$.  2) Prove, that $\operatorname{Inn}(G) = \{p_a \mid a \in G\}$ is a normal subgroup of $(\operatorname{Aut}(G),\circ)$.
54131
Let $G$ be a group. Let $n$ be the number of subgroups of $G$ that are not normal. Prove that $n\not=1$.     When $G$ is Abelian, then this is trivial since $n=0$. For the non-Abelian group $G$, I have to show that $n\geq 2$. But I don't see how to go on. Any hint?