A and B are to nonempty subset of a group G with $|A|+|B| &gt; |G|$ then G = AB.  I am a bit confused of the order of a subset of a group. Generally a subset of a group is not a subgroup but when it occurs $|A|+|B| &gt; |G|$, I want to figure out the relation between A and B and if A will have the inverse of some elements of A if $|A| &gt; \frac{|G|}{2}$?
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.
76923
Suppose that a set has an odd number of elements. Explain why half of the subsets will have an odd number of elements .  Now assuming set A is the set with an odd number of integer elements{1,2,3,4,5}  Subset b ={1,2}  subset c={3}  Subset d={4,5}  Now there are two subsets with even number of elements and one subset with an odd number of elements.It seems to contradict the earlier theorem that i have to prove.I think that i am missing something.
88763
Let G be a group with $o(G)=pq$, $p&gt;q$ are primes.  Prove that $G$ has a subgroup of order $p$ and a subgroup of order $q$.  Remark: I cannot find the topic with this problem.  I have tried different methods but no one of them did not bring result. Please show how to do that. Do not use Sylow's Theorem.
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Studying groups and subgroups I find this question:     Are there subgroups of order $\mathbf 6 \mathbf 5$ in the additive group $(\Bbb Z$,$+)$?   I would answer no, because a subgroups of $(\Bbb Z,+)$ is the multiple of a Natural number $n $ and it has the form: $n\Bbb Z$={$na|n \in \Bbb N, a \in \Bbb Z$} and they have no finite order.  But I'm not sure of this answer.  Could someone explain if I'm wrong and why?  Thanks.  $\mathbf {edit}:$  My question comes because discussing with a course mate he argued that exist Z/65Z in (Z,+), the group module 65 and it  has order 65 so I'm a bit confused.
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What does it mean when you have a set like A = {1, {2}, 3, 4}? Is {2} a subset within A?  Say I have the following:  A = {1, {2}, 3, 4}, B = {1, 2, {3}}, C = {1, 2}  How many elements are in $A \cap B \cap C$? 1 or 2?  What if I want to find $A \times B$, is there 6 elements or 12?
335
Let $G$ be a group with $|G| = n$ and let $ \emptyset \ne S \subseteq G$.   I want to show that $S^n$ is a subgroup of $G$ where by $S^n$ I mean the set $\lbrace s_1\cdots s_n \; | \; s_i \in S\rbrace$.
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I came to point where I suppose for case 1 that $A\subseteq B$ and conclusion is trivial. For case 2 I suppose that $A\not\subseteq B$ and try to prove $B\subseteq A$, but that gets me nowhere. Any pointers here are most welcome.
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What is the intuition behind the fact that if every element in a group is of order $2$, we have that the group is abelian? I can prove it, but I do not know the intuition behind it.
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Let $G$ be a group and $c$ be a fixed element of $G$. Define a new operation $*$ on G by:   $$x*y=xdy$$   for all $x$ and $y$ in $G$, where $d$ is the inverse of $c$. Prove that $G$ is a group under the operation $*$.    If you want to check, this problem is in the book “A Course In Group Theory”, by John F. Humphreys (problem $3$, page $28$)   I know what are the axioms that $(G,*)$ must obey in order to be a group and I did show that all the axioms where obeyed except the inverse axiom.  My problem is that, whenever I try to verify what would the inverse be for some element $x$, I get an equation where the $x$ is cancelled in both sides.  Can please someone guide me?  Thank you!  (P.S.: This is NOT a homework assignment, I’m studying Group Theory by this book and got problems solving this question)
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.
32357
Let $G$ be a finite group. Suppose that $A$ and $B$ are to subsets of $G$. If $|G|&lt;|A|+|B|$ prove that $$G=AB.$$
30957
In a question from an exam in an undergraduate group theory course, we were asked to prove or disprove that the set of all Torsion elements of a group is necessarily a subgroup.  I knew that the set of Torsion elements is closed under the inverse operation, but was later told that it is not closed under multiplication, therefore disproving the claim. However, I couldn't find any examples of a group $G$ and two elements $a,b$ such that both $a$ and $b$ have a finite order, but $ab$ doesn't. I know that for this to happen $G$ must be an infinite and non-Abelian group, but still couldn't find a valid example.   What are some examples of groups/elements fulfilling the aforementioned property?
1071
Let $G$ be a group and let $a,b$ be two elements of $G$. What can we say about the order of their product $ab$?   says "not much":     There is no general formula relating the order of a product $ab$ to the orders of $a$ and $b$. In fact, it is possible that both $a$ and $b$ have finite order while $ab$ has infinite order, or that both $a$ and $b$ have infinite order while $ab$ has finite order.   On the other hand no examples are provided. $(\mathbb{Z},+), 1$ and $-1$ give an example of elements of infinite order with product of finite order. I can't think of any example of the other kind! So:   What's an example of a group $G$ and two elements $a,b$ both of finite order such that their product has infinite order?     Wikipedia then states:     If $ab = ba$, we can at least say that $\mathrm{ord}(ab)$ divides $\mathrm{lcm}(\mathrm{ord}(a), \mathrm{ord}(b))$   which is easy to prove, but not very effective. So:   What are some similar results about the order of a product, perhaps with some additional hypotheses?
77527
(Reference : Fraleigh, A first course in abstract algebra)     Prove that a nonempty set $G$, together with an associative binary operation * on $G$ such that $a*x=b$ and $y*a=b$ have solutions in G, $\forall a,b\in G$, is a group.   Because * is an associatie operation, we have to prove that: there exists $e\in G$ such that $e*a=a=a*e$ for any $a\in G$, and that for any $a\in G$ there exists $\alpha \in G$ such that $a*\alpha=e=\alpha *a$.  This is very confusing, first I tried to prove the identity exists. If we consider a=b, then $a*x=a=y*a$, and because of the hypothesis $x,y\in G$ always exists, now we have to show that $x=y$, but I just can't do this with out inverses, so I tried to prove that the inverses exists, however I have no idea how the identity looks like, I'm stuck.  And I tried to get unstuck, so I thought particular cases, first if $G$ had only one element, then what would happend if it had just two elements... but I remembered that we don't know if G is finite or not, and I think this makes things a little bit more complicated.  What can I do?
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.
24118
If the order of all nontrivial elements in a group is 2, then the group is Abelian. I know of a proof that is just from calculations (see below). I'm wondering if there is any theory or motivation behind this fact. Perhaps to do with commutators?    Proof: $a \cdot b = (b \cdot b)  \cdot (a \cdot b) \cdot (a \cdot a) = b \cdot (b \cdot a) \cdot (b\cdot a) \cdot a = b \cdot a$.
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Suppose $G$ is a finite group and I know for every $k \leq |G|$ that exactly $n_k$ elements in $G$ have order $k$. Do I know what the group is? Is there a counterexample where two groups $G$ and $H$ have the same number of elements for each order, but $G$ is not isomorphic to $H$? I suspect that there is, but I haven't thought of one.