Let $H \subset G$ a subset non-empty of finite cardinality of a group   $(G,\circ)$ such that $h_1 \circ h_2 \in H$ for each $h_1,h_2 \in H$.   Then $(H,\circ)$ is a subgroup of $(G,\circ)$.   In order to show that it is a group, I already know that $h \circ 1_G = h$, for $h \in H$, but I am not able to bring the $1_G$ in our subset $H$.  I don't know how to use it that our subset cardinality is finite and non-empty. Does someone could help me on this?
121383
I have some idea, lets take any element $a$ from that subgroup and repeatedly apply group operator over it, since group is finite that element $a$ will appear again after some time, $\{a, a^2, a^3,...,a\}.$  Then element previous to 2nd $a$ should be identity element. And prior to it is $a$'s inverse. Hence it has all property of groups. Is this proof correct.  I am not sure about that can 2nd $a$ appear only if we have identity element in it. Please help.  Reference: Fraleigh p.58  Question 5.50 in A First Course in Abstract Algebra
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?
78057
Let $G$ be a finite group, say a group with $n$ elements, and let $\emptyset \neq S \subseteq G$. Suppose $e \in S$, and that $S$ is closed with respect to multiplication. Prove that $S$ is a subgroup of $G$.    Let $G = \{a_1, \ldots, a_n\}.$ Let $a_i \in G.$ Then any number of $a_ia_1, a_ia_2, \ldots, a_ia_n \in S,$ since $S$ is closed under multiplication. One of those $a_ia_j$ is $e$ since $e \in S.$ Then either $a_i$ or $a_j$ is $a^{-1}.$ Would this argument work?
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.
85706
Let $G$ be a group and $N$ a subgroup of $G$. I read that the following two definitions are equivalent:  1) $\forall g\in G$, $gNg^{-1}=N$  2)$\forall g\in G$, $gNg^{-1}\subset N$  does this mean that we have always $N\subset gNg^{-1}$ ? my guess is no because take $n\in N$ then if $n=gn&#39;g^{-1}$ then $n&#39;=g^{-1}ng$ and there is no reason that $g^{-1}ng\in N$
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 $k'\in K$, $gk'g^{-1}=k\in K\Rightarrow k'=g^{-1}kg\in\{gkg^{-1}\mid k\in K\}$.
114819
If $G$ is a group, $H$ is a subgroup of $G$ and $g\in G$, is it possible that $gHg^{-1} \subset H$ ?    This means, $gHg^{-1}$ is a proper subgroup of $H$. We know that $H \cong gHg^{-1}$, so if $H$ is finite then we have a contradiction since the isomorphism between the two subgroups implies that they have the same order so $gHg^{-1}$ can't be proper subgroup of $H$.  So, what if $H$ is infinite is there an example for such $G , H , g$ ?   Edit: I suppose that $H$ has a subgroup $N$ such that $N$ is a normal subgroup of $G$.
73465
I have been going through some material on elementary group theory () and in one place it states:  A subgroup which contains half of all elements, i.e. for which $|G|=2|H|$, is a normal subgroup.  Where a subgroup $H\subset G$ is called a normal subgroup if $gHg^{-1}=H~\forall g\in G$. How would I go about proving this? Sorry if the answer exists somewhere, but I tried searching and could not find it.
58373
I want to show that if $G$ is a group with more than one element, and that $G$ has no proper non-trivial subgroups. Prove that $|G|$ is prime. (Do not assume at the outset that |G| is finite).  My question is not that how to prove it.  I am saying that suppose $|G|\geq 2$ possibly $|G|=\infty.$ By assumption the only subgroups of $G$ are $\{e\}$ and $G$, i.e., the trivial groups. Let $a$ be non-identity element in $G$. Consider $\langle a\rangle$. Then  $\langle a\rangle=G.$ So $G$ is cyclic.   My question is, why can I say that $G=\langle a\rangle$. I know there are only two subgroups and $\langle a\rangle\neq e$ because $a\neq e$. Therefore we must have $G=\langle a\rangle$. But my problem is why cant I say that consider $a,b\in G$ and then we look at $\langle a,b\rangle$. And then I say $G=\langle a,b\rangle$ and then I cannot say that $G$ is cyclic, and then I will have problem proving question.
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.
85423
If H is a subgroup of G ( | G:H |=n, n > 1), then how can you show that G has a factor group G/N which is isomorphic to a subgroup D of the symmetric group Sn with |D| > 1.  I've been looking at Cayley's Theorem but it's not specific to the factor groups.
33139
Let $G$ be a finite and non-abelian group.  How do I prove the following statement? $$\exists a,g,h\in G \colon\quad h=aga^{-1},\ g\neq h ,\ gh=hg.$$  Thanks in advance.
39758
Prove the following: If $H$ is a subgroup of finite index in a group $G$, and $K$ is a subgroup of $G$ containing $H$, then $K$ is of finite index in $G$ and $[G:H] = [G:K][K:H]$.  So this is basically a bijective proof? The number of cosets of $H$ in $G$ equals the number of cosets of $K$ in $G$ times the number of cosets of $H$ in $K$ by the multiplication principle?   Reference: Fraleigh p. 103  Question 10.35 in A First Course in Abstract Algebra
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 $g = hk$)
80653
Let $G$ be a group and $\{H_i\}_{i\in I}$ a family of subgroups. I would like to find a condition that will imply that $\cup_{i\in I} H_i$ is a subgroup. I know that it's not true in general, I need help to find this condition.  Any help is welcome.  Thanks a lot.
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.
120795
Let G be a finite group, and let S and T be (not necessarily distinct) nonempty subsets. prove that either $G=ST$ or |$G|\geq|S|+|T|$  That's my thougt, I am thinking suppose $G$ does not equal to $ST$, then there exist a $h$ in $G$, but $h$ $\notin ST$. Thus, I suppose T={$x_1,x_2......x_N$}and $S=\{{t_1,t_2,.......,t_m,x_{k+1},......,x_{N}}\}$, where $x_k+1,...x_N$ are the common elements between two sets S and T. So, I want to construct two totally different sets A and B so that $|A|+ |B| &gt; |G|$, firstlt,if $1$ is not in $S$ and $T$, then suppose T=A, and then suppoese B={$h,t_1,...,t_m, hx_{k+1}^{-1},...,hx_N^{-1}$}. I think if we can prove B$\neq$A(orT) and B$\neq$S, then I can know the question will be proved since $|A|+|B|&gt;|G|$, but I only can prove B$\neq$S, I don't know how to prove B$\neq$T. Maybe my thoughts are wrong, can someone tell me how to solve this question? I think my methods are not correct, I hope someone can help me solve this question.