Claim : $H$ and $K$ are two normal subgroups of group $G$ such that $H\cap K = \{e\}$ then $ xy = yx $ for $x \in H$ and $y \in K$.  Proof : Let $x \in H$ and $y\in K$ then I need to prove that $xy =yx$. Let us assume that   $$y^{-1}xy \neq x$$ $y^{-1}xy = z$ where $z \in H$  Question : I am not getting how to proceed further?
106873
This is the proof, which I mostly understand except for one bit:  You have $h_1 \in H_1$ and $h_2 \in H_2$.   We also have $h_1^{-1}(h_2^{-1}h_1h_2) \in H_1$, because $h_2^{-1}h_1h_2 \in h_2^{-1}H_1h_2 = H_1$. Similarly, we have $(h_1^{-1}h_2^{-1}h_1)h_2 \in H_2$. Therefore  $$ h_1^{-1}h_2^{-1}h_1h_2 \in H_1 \cap H_2 = \{1_G\} $$  and so $h_1^{-1}h_2^{-1}h_1h_2 = \{1_G\}$. Let's first multiply everything on the left by $h_1$  $$h_1^{-1}h_2^{-1}h_1h_2 = \{1_G\}$$ $$ h_1 h_1^{-1}h_2^{-1}h_1h_2 = h_1 \{1_G\}$$ $$ h_2^{-1}h_1h_2 = h_1 \{1_G\}$$   Multiply both sides on the left by $h_2$ giving us  $$ h_1h_2 = h_2 h_1 $$  The bit I don't get is right at the beginning. Why is this correct: "We also have $h_1^{-1}(h_2^{-1}h_1h_2) \in H_1$, because $h_2^{-1}h_1h_2 \in h_2^{-1}H_1h_2 = H_1$."
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?
43020
I am wondering how to prove the formula $|HK| = \dfrac{|H| \; |K|}{|H \cap K|}$ where $H$ and $K$ are finite groups. Let $|H|$ denote the order of $H$ and so on.  I have seen proofs from these websites:  and .  But I do not really understand the idea of the proof. Can someone explain the proof to me, please? Thanks so much.
75242
Let $H$ and $K$ be subgroups of a finite group $G$.  A standard result is the formula $$\left|HK\right| = \frac{\left|H\right|\left|K\right|}{\left|H \cap K\right|}$$ for the cardinality of the set $HK$. The proofs of this that I've seen always involve some arbitrary choice, such as the choice of a transversal for $\{hK \colon  h \in H\}$. Is there a way to show that these quantities are equal without making such an arbitrary choice?  For example, can they be shown to be the respective dimensions of a pair of canonically isomorphic modules?
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.
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.
18696
Let $G_1$ and $G_2$ be finite groups and let $K \le G_1 \times G_2$      (i) Set $H_1 = \{ g \in G_1 : (g,e) \in K \}$ and $H_2 = \{ g \in G_2 : (e,g) \in K \}$. Show that:   $$H_1 \le G_1; \quad H_2 \le G_2; \quad H_1 \times H_2 \le K.$$      (ii) Suppose that $|G_1|$ and $|G_2|$ are coprime. Show that $K = H_1 \times H_2$.      (iii) Show that this result need not follow if $|G_1|$ and $|G_2|$ are not coprime.   I have done part (i). I was doing part (ii) and got stuck:  Since from above I showed that $H_1 \times H_2 \subseteq K$, now I only need to show that $H_1 \times H_2 \supseteq K$.  Let $(g_1,g_2) \in K$ where $g_1 \in G_1$ and $g_2 \in G_2$. Since $K \leq G_1 \times G_2$, we have $(g_1,e) \in K$ and $(e,g_2) \in K$. Hence $g_1 \in H_1$ and $g_2 \in H_2$.  Finally we have $(g_1, g_2) \in H_1 \times H_2$  As I haven't used the fact that order of G_1 and G_2 are coprime, there must be something fundamentally wrong with my proof.
32086
Let $G$ be any group, and $H \leq G$ a subgroup.  Suppose that for each $x \in G$, there exists a $y \in G$ such that $xH \subseteq Hy$. In other words, every left coset of $H$ is contained inside some right coset of $H$.  Question: what can we say about $H$? In particular, does this imply that $H$ is normal?  I know that if $G$ is finite, then $H$ must be normal, because then $xH \subseteq Hy$ implies $xH = Hy$, since $|xH| = |Hy|$.
73475
I have a group $G=\mathbb{Z}_{36}$ with normal subgroups $K$ and $H$, where $H$ is also a normal subgroup of $K$. What I'm really supposed to find is a bijection between $G/K$ and $(G/H)/(K/H)$ by using the third isomorphism theorem. I have already calculated $G/K$, $G/H$ and $K/H$, but I do not really know how to calculate $(G/H)/(K/H)$. Say that one of the elements in $G/H$ is $\{0,18\}$ and one of the elements in $K/H$ is $\{0,18\}$. Will one of the elements in $(G/H)/(K/H)$ be as follows? $$\{0,18\}+\{0,18\} = \{0+0,18+18\}=\{0,36\}=\{0\}$$ Another way I'm thinking it's possible to caluclate is $$\{0,18\}+\{0,18\} = \{0+0,0+18,18+0,18+18\}=\{0,18,18,36\}=\{0,18\}$$ Is any of these a correct way of calculating an element in $(G/H)/(K/H)$?
115328
Let $G$ be a group and $n\geq 2$ an integer. Let $H_1$ and $H_2$ be two distinct subgroups of $G$ that satisfy $$[G:H_1]=[G:H_2]=n\text{ and } [G:H_1\cap H_2]=n(n-1).$$ I've been asked to prove that $H_1$ and $H_2$ are conjugate. My thoughts This is the same as saying that some coset of $H_1$ coincides with some coset of $H_2$, so I think the argument must be in the line of concluding that there cannot be so many different cosets. From the index theorem I've concluded that $[H_1: H_1\cap H_2]=[H_2: H_1\cap H_2]=n-1$ but I don't know if this is useful. I've thought of considering all possible intersections between a coset of $H_1$ and a coset of $H_2$. There are $n^2$ such intersections and only $n(n-1)$ cosets of $H_1\cap H_2$. By  I know that these intersections are either empty or a coset of the intersection, so there are exactly $n$ intersections that are either empty or repeated. I have no idea how to continue from this, any help would be appreciated. EDIT: I've also tried to treat the particular case $H_1\cap H_2=1$, in which $[G:H_1\cap H_2]=|G|=n(n-1)$, but I don't really see any shortcut in this case.
54096
Is the following proof correct, or are certain steps illegal?  Suppose $f:G_1\to G_2$ is a surjective group isomorphism, $N_1\lhd G_1, N_2\lhd G_2$ normal subgroups, with $N_1=f^{-1}(N_2)$. Prove that $(G_1/N_1)\cong (G_2/N_2)$.  Proof:  As $f$ is surjective, we an use the first isomorphism theorem to get $G_1/\ker(f)\cong G_2$, which implies that $$(G_1/\ker f)/(N_1/\ker f)\cong G_2/(N_1/\ker f)$$  Using the first isomorphism theorem we also know that $$N_1/\ker (f)\cong f(N_1)=f(f^{-1}(N_2))=N_2$$  Combining this we can see that $(G_1/\ker f)/(N_1/\ker f)\cong G_2/(N_1/\ker f)\cong G_2/N_2$.  Using the third isomorphism theorem we have that: $$ (G_1/\ker f)/(N_1/\ker f)\cong G_1/N_1$$  And from the last two lines we have $G_1/N_1\cong G_2/N_2$ $\tag*{$\Box$}$
50167
Let $K$, $N$ be normal subgroups in $G$.  I would like to show that if $K\cap N=\{e\}$, then $KN\cong K\times N$.  My original thought was to invoke the first Isomorphism theorem.  Therefore I would have to find an appropriate mapping $\varphi$ whose kernel is $\{e\}$.    In a previous exercise, I showed, again by the first Isomorphism theorem that   $$\frac{KN}{K\cap N}\cong\frac{KN}{N}\times\frac{KN}{K}$$  thus, if $K\cap N=\{e\}, $ i would just have to show that $\frac{KN}{N}=K$ and $\frac{KN}{K}=N$.  Thus my idea was to prove that under the condition that the intersection of $K$ and $N$ that elements in $KN$ can commute freely.  We know that $KN=NK$, but this does not imply commutativity, only that if $kn\in KN$, then $kn=n'k'$ with $n$ not necessarily equal to $n'$ and such with $k$.  I just can't come up with a commutativity proof.  EDIT: I forgot to define $KN=\{kn|k\in K, n\in N\}$.
93319
Let $G=GL(2,\mathbb R)\oplus GL(2,\mathbb R)$ and let $H=\{(A,B)\in G\mid \det(A)=\det(B)\}$. Prove that $G/H \simeq (\mathbb R^*,\times)$.   I'm guessing I should use: "Let $G$ be a group and let $H$ be a normal subgroup of $G$. The set $G/H = \{aH\mid a \in G\}$ is a group under the operation $(aH)(bH)=abH$."
61365
Let $G$ be any group. It's a well-known result that if $H, K$ are subgroups of $G$, then $HK$ is a subgroup itself if and only if $HK = KH$.  Now, I've always wondered about a generalization of this result, something along the lines of:  Theorem: If $H_1, \ldots, H_n$ are subgroups of $G$, then $H_1H_2\dots{H_n}$ is a subgroup if and only if ($\star$) holds, where $(\star)$ is some condition on $H_1, \ldots, H_n$, preferably related to how the smaller products $H_{m_1}\ldots{}H_{m_k}$, for $k &lt; n$, behave.  Question 1: Is there such a theorem?  I do know, and its easy to prove, that if $H_iH_j = H_jH_i$ for every $i, j$, then the big product is a group, but this is not satisfying since it's far from necessary (just take one of the groups to be $G$, and you need no commutativity at all). Also, I've been told that there is no really satisfactory answer; if that is indeed the case, then my question would be why? In particular:  Question 2: Are there really problematic counterexamples where you can see that the behavior of the smaller products has nothing to do with the big product, so that no such a theorem can ever exist?  I would appreciate even an answer for the particular case $n = 3$.  Thanks.
85692
I am trying to prove the following statement:     Let $G$ be a group and $H, K\mathrel{\unlhd}G$. Assuming that $H \cap K =  \{1_G\}$ and $G = \langle H, K \rangle$, prove that $G \cong H \times K$.   From this, we get that $HK \cong H \times K$. Naturally, it would be ideal if I could prove that $HK \cong G$. One issue that I am facing here is the meaning of $G = \langle H, K \rangle$. How is this defined (I mean, more explicitly)? Is my strategy correct or
52402
Let $G$ be a finite Group, let S=$\{s_1,s_2,s_3 ... |s_i\in G $ has odd order $\}$. Let $H=\langle S\rangle$ The question is:   Prove that $H$ is normal subgroup of $G$. prove that $G/H$ has order $2^k$ for $k \in \mathbb{N}$.   I tried look at $$gHg^{-1}$$ but can't understand why $gHg^{-1} \subseteq H$ probably because I cant understand what unique in H. Thanks much for help.
6172
$H$ is subgroup of $G$ with $H$ not equal $G$.  Be $S=G-H$. I am being asked to prove that $\langle S \rangle=G$.  Some tip to solve this? I think in $S_3$ is possible but I can´t prove.