Suppose G is a group, with subgroups H and K.   Prove that if H ∪ K is a subgroup of G implies that H ⊆ K or K ⊆ H.   I'm not really sure how to start this, I can prove that H ∩ K is a subgroup but I don't know how to approach the union. Any help is appreciated.
123438
Let $H$ and $K$ denote two subgroups of a group $G$.  Prove that the union $H \cup K$ is a subgroup of $G$ if and only if $H \subset K$ or $K \subset H$.
34191
Let $H$ be a subgroup of $G$. Prove that $gHg^{-1}=\{ghg^{-1}|h \in H\}$ is a subgroup of $G$, $g\in G$. Show that $o(gHg^{-1})=o(H)$.   Attempt:  As $e=geg^{-1}\in gHg^{-1}$ then $gHg^{-1}$ is non-empty. Let $x=gh_1g^{-1}$, $y=gh_2g^{-1}\in gHg^{-1}$, $h_1, h_2\in H$   then $xy^{-1}=\cdots=g(h_1h_2^{-1})g^{-1}\in gHg^{-1}$ as $h_1h_2^{-1}\in H$ ($H$ being the subgroup of $G$)  Therefore, $gHg^{-1}$ is a subgroup.   The problem is in 2nd part. I am unable to sgow that $o(gHg^{-1})=o(H)$. Please help me to solve the problem using the elementary tools of abstract algebra.
88559
Let H be a subgroup in a group $G$. If $G$ is generated by a set $S$, so $G=\langle S\rangle $, prove that $H$ is a normal subgroup in $G$ iff $sHs^{-1}$ $\subseteq H$ for all $s\in S$.    Note with the question: Results of this sort are useful because it is easier to check an algebraic property for a small set of generators than to prove it holds for all elements of G.  Hi, the forward direction is quite straightforward but I will be really grateful if somebody can help me with the reverse direction. Thanks!
38710
Let $G$ be a group and $H$ a subgroup of $G$. For $x \in G$, prove $x^{-1}Hx$ is a subgroup of $G$.
50299
Prove that if $G$ is abelian then the set $H$ of all elements of $G$ that are their own inverses is a subgroup of $G$.   Naturally in an abelian group, $ab = ba$ for $a, b \in G$, however I'm not sure how to show the set elements that are their own inverses is a subgroup of $G$ using arbitrary elements.
10989
Question: Let $G$ be a group. Prove that $Z(G)$ is a subgroup of $G$.   If I want to show that $Z(G)$ is a subgroup of $G$ that means I have to show that it is closed under group operation?  Here is my attempt.   Let $a,b$ be elements in $Z(G)$ and $x$ be an element in $G$. Then $ax=xa$ which is under group multiplication commutative and under inverse $(a^{-1}) x=x(a^{-1})$. And hence $(a^{-1})bx= (a^{-1})xb=x (a^{-1})b$ which is under group operation so $(a^{-1}),b$ are elements in $Z(G)$ thus a subgroup of $G$.  Really appreciate if anyone can help me by directing me if my attempt is not good. Thanks in advance.
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 $k'\in K$, $gk'g^{-1}=k\in K\Rightarrow k'=g^{-1}kg\in\{gkg^{-1}\mid k\in K\}$.
44913
This is , I think an easy problem just that I am not getting the catch of it. How to show whether or not the statement is true?  All subgroups of a group are normal$\implies$ the group is an abelian group?  I have been able to show the other way round.
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
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?
93764
Let n be a positive integer and let G be an Abelian group. Define  $H := \{x ∈ G : |x| \ \text{divides} \ n\}$  where x denotes the order of |x|.  Show that H is a subgroup of G.  I know I need to prove that H is non-empty, contains the identity, e, and that it contains $ab^{-1}$.  I'm just not sure how to prove this, thank you for any help!
10179
Let $G$ be a group. We define $H=\{h\in G\mid \forall g\in G: hg=gh\},$ the center of $G$. Prove that $H$ is a (normal) subgroup of $G$.
18013
I am participating in an Introductory course in groups and I have the following question:     Let $G$ be a finite group of order $8$. Prove that $G$ has a subgroup of order $4$ and a subgroup of order $2$.   i know how to prove that $G$ has a subgroup of order $2$ but i can't show the existence of a subgroup of order 4. I am only a begginer in groups so please take this into account.
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.
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 $G$.
11823
I'm having some trouble understanding cosets.  If I understand it correctly, cosets form a partition. So, if I have $|G:H| = 2$, then $G = H \cup xH$. Right?  In an exercise, I'm asked the following:  "For $G$ group and $H \le G$ such that $|G:H| = 2$, prove that $x^2 \in H$ for any $x \in G$"  If $x \in H$, then $x^2 \in H$ because $H$ is a group. But, if $x \notin H$, then $x \in gH$ for any $g \in G$, right? Then $x^2$ = $(gh)^2$ but I can't see why this is in $H$.  Can you give me some hint? Thanks for stopping by.
30868
Just out of curiosity: If $G$ is a topological group and $H, K$ are closed subgroups, is $H\cdot K$ a closed subgroup?  Thanks!