This is an exercise from Hungerford chapter 2 section 3. Here $G $ is said to satisfy the ACC and DCC. Then how should I apply the Krull Schmidt theorem to prove the proposition in the title? I tried to show that $G×G $ itself satisfies the ACC and DCC but to no avail. Could anyone help me?
123633
I was thinking about the following problem:  Suppose that $G_1 \cong G_2$ are isomorphic groups. Under what conditions on the groups $H_1,H_2$ will we have $$G_1 \times H_1 \cong G_2 \times H_2 ?$$  Obviously, in the finite case we must have $|H_1|=|H_2|$. Also, it is easy to see that $H_1 \cong H_2$ is sufficient. Is it necessary as well?  Thank you!
103317
Problem : Let $G$ be a p-group and $G'$ denote the commutator subgroup of $G$. If $[G' : G'']\leq p^2$, then $G'$ is abelian.  It is easy to prove it for the case of $[G' : G'']=1$ since G is solvable. But it is too hard to prove the other cases. Please give me any advice or solutions. Thanks in advance.
109254
What I want to ask is the proof of the Proposition 10. in "Linear Representations of Finite Groups" by Jean-Pierre Serre.   Let $\rho_i : G \rightarrow GL(W_i)$ be the distinct irreducible representations of $G$.$(1 \le i \le h)$ We can extend $\rho_i$ to $\widetilde{\rho_i} : \mathbb{C}[G] \rightarrow \text{End}(W_i)$ by $\widetilde{\rho_i}(\sum_{g \in G} a_g g) = a_g \sum_{g \in G} \rho_i(g)$.   Then $\widetilde{\rho} = (\widetilde{\rho_i}) : \mathbb{C}[G] \rightarrow \prod_{i=1}^{h} \text{End}(W_i) \cong \prod_{i=1}^h M_{n_i}(\mathbb{C})$ is a homomorphism.  Then Proposition 10 states that it is an isomorphism.  Since dimensions of domain and codomain are same, it suffices to show that $\widetilde{\rho}$ is surjective.   In the proof of the book,   If $\widetilde{\rho}$ is not surjective, then there exists a nonzero linear form on $\prod M_{n_i}(\mathbb{C})$ vanishing on the image of $\widetilde{\rho}$. This gives a nontrivial relation on the coefficients of the representations $\rho_i$ which is impossible because of the orthogonality formulas of 2.2.  But I can not follow the above three lines in the proof. I will appreciate it if you give an explanation for the proof.   Thank you.
42847
Let $G$ be a finite group. Let $N \trianglelefteq G$ and $U \leq G$ such that $G = NU$. Then there exists a bijection, preserving inclusion, from the set of subgroups $X$ satisfying $U ≤ X ≤ G$ to the set of $U-$invariant subgroups $Y$ satisfying $U \cap N ≤ Y ≤ N$.  This is a problem from Kurzweil-Stellmacher book. I dont know how to do it. Thanks for any help.
69924
This theorem is obviously correct. Now I try to prove it by well-ordering principle.  But I don't know where to start the proving....
49140
See the related question . This is the second part of question 4-C in Milnor and Stasheff's book on characteristic classes. In the solution to the first part, we rely on the fact that having a field of tangent $1$-planes allows us to split the tangent bundle into a direct sum. It doesn't seem like this should be true for a field of tangent $2$-planes (otherwise the question wouldn't just be asking about $P^4$ and $P^6$), so I'm not sure exactly where to start.  edit: actually, since $P^n$ can be given a Riemannian metric, any subbundle should be a Whitney summand; this doesn't answer the question as to why they only ask for $P^4$ and $P^6$ though, so I'm still confused.
66002
How would you show that if $ac≡bc$ $\mod m$ and $\gcd(c,m)=d$, then  $a≡b$ $\mod \frac{m}{d}$?  Any help would be much appreciated!
61307
Since $T_p G$ is isomorphic to $T_e G$ for all $p\in G$, it makes sense that each vector in $T_p G$ can be identified with a vector in $T_e G$. Hence, to make the map from $TG$ one to one, we must identify each vector in $T_e G$ with a point in $G$, but I'm having trouble saying this precisely.  Thanks so much!
115635
Let $c(G)$ denote the number of conjugacy classes of a group $G$. I have to show that  $$\lim_{n \to \infty} \inf _{|G|=n}c(G)=\infty.$$ That is, I have to show that $\exists $ a function $f:\mathbb{N} \to \mathbb{N}$ such that if $c(g)\leq n, $ then $|G|\leq f(n).$   I know that if $G$ is a finite group such that any two non-identity elements in $G$ are conjugate, then $G$ is isomorphic to $\mathbb{Z}_2$. The proof is simple, infact:      It can easily be shown that $G$ acts transitively by conjugation on $G \setminus \{e\}$. Then, by  orbit-stabilizer theorem, the claim follows.    I wonder if similar idea could be useful somehow. But I can't think of any to prove the claim above. Please help!
94063
The following was the B6 problem of 1985 Putnam Competition: Suppose $G$ is a finite group (under matrix multiplication) of real $n\times n$ matrices $\{M_i\}, 1\leq i\leq r$. Suppose that $$\sum_{i=1}^r tr (M_i)=0$$, prove that $$\sum_{i=1}^rM_i=0.$$  Here is an official proof from the committee which I didn't understand:  Lemma: Let $G$ be a finite group of order $r$. Let $\rho: G\rightarrow GL(V)$ be a representation of $G$ on some finite dimensional vector space $V$. Then $$\sum_{g\in G}tr \rho_g$$ is a non-negative integer divisible by $r$, and is zero iff $$\sum_{g\in G}\rho_g=0$$.  Proof: Let $\chi_1,\cdots, \chi_s$ be the irreducible characters of $G$ and $\chi= \sum_{i=1}^s a_i\chi_i$ and $\psi=\sum_{i=1}^sb_i\chi_i$ be arbitrary characters. Then by the orthogonality relations of characters, we have $$\frac{1}{|G|}\sum_{g\in G}\chi(g)\overline{\psi (g)}=\sum_{i=1}^sa_ib_i$$. Applying this to the character of $\rho$ and the trivial character $\mathbb{1}$ shows that $\frac{1}{|G|} \sum_{g\in G}tr \rho_g$ equals the multiplicity of $\mathbb{1}$ in $\rho$, which is a non-negative integer. Now suppose that the matrix $S=\sum_{g\in G}\rho_g$ is non-zero. Choose $v\in V$ with $Sv\not=0$. The relation $\rho_hS=S$ shows that $Sv$ is fixed by $\rho_h$ for all $h\in G$. In other words, $Sv$ spans a trivial subrepresentation of $\rho$, so the non-negative integer of the previous paragraph is positive.  QED  We now return to the problem at hand. "Unfortunately the $M_i$ do not necessarily define a representation of $G$, since the $M_i$ need not be invertible." Instead we need to apply the lemma to the action of $G$ on $\mathbb{C}^n/K$ for some subspace $K$ ...  I do not understand wthe sentence in "". Isn't the set of $M_i$'s form  a group under multiplication? why they need not to be invertible? The above proof is copied from Kedlaya, Poonen and Vakil's Putnam competition 1985-2000. Thanks for helping
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 $g = hk$)
44684
Let $R$ be a UFD. Then for any two elements $a,b \in R \setminus \{0\}$ $ab \sim \gcd (a,b){\operatorname{lcm} (a,b)}$.   My attempt :  Let $d = \gcd (a,b)$.Then $d|a$ and $d|b$.Then $\exists x,y \in R$ such that $a = dx$ and $b = dy$.Now it is to be shown that $c = dxy$ is the lcm of $a,b$.Which I find difficulty to show.It is clear that $a|c$ and $b|c$. Now if $u$ is any common multiple of $a$ and $b$.Then $dx|u$ and also since $y|b$ we have $y|u$ i.e. $dxy|u^2$ $\implies c|u^2$.But I have to show $c|u$ to complete the proof which I fail to prove.Please help me to complete the proof.  Thank you in advance.
4942
Suppose $f:X\rightarrow Y$ is a surjection and you want to show that there exists $g:Y\rightarrow X$ s.t. $f\circ g=\mathrm{id}_Y$. You need the AC to show this.  However, suppose $f$ is a injection and you want show there is $g$ s.t. $g\circ f=\mathrm{id}_X$. Then, according to my textbook, you don't need the AC to show this.  This is counterintuitive to me, because it's like you need a special axiom to claim that an infinite product of big sets is nonempty, while you don't need one to claim that an infinite product of singleton sets is nonempty, which seems smaller than the former.  So why don't you need the AC to show the latter?  EDIT: $X$ should be nonempty.  EDIT 2: I realized (after asking this) that my question mostly concerns whether the AC is needed to say that an infinite product of finite sets is nonempty, and why.
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.
46374
From :     Importantly, the DDH assumption does not hold in the multiplicative group $\mathbb {Z} _{p}^{*}$, where $p$ is prime. This is because given $g^{a}$ and $g^{b}$, one can efficiently compute the Legendre symbol of $g^{ab}$, giving a successful method to distinguish $g^{ab}$ from a random group element.   The only way I can think of to distinguish $g^c$ from $g^{ab}$ is:    If $\left(\frac{g}{p}\right)=-1$, then $\left(\frac{g^{ab}}{p}\right) = 1$ if and only if $a$ or $b$ is even (which is $3/4$ of the times). So if $\left(\frac{g^c}{p}\right) =-1$, probably $g^c\neq g^{ab}$. Nothing is certain.   How'd do you compute $\left(\frac{g^{ab}}{p}\right)$ given $g^a$ and $g^b$?
77555
Let $\langle g\rangle$ be a cyclic group of order $n$. Suppose $1\leq q \leq n-1$, I want to show that $g^q$ generates $\langle g\rangle$ if and only if $\gcd(n,q)=1$.  Suppose $g^q$ generates $\langle g\rangle$, then   $$ 1,g^q,g^{2q},...,g^{(n-1)q} $$  are all distinct and $g^{nq}=1$, since $n$ is the order of the group. Note that this is the lowest multiple of $q$ for which this is the case. Therefore $\operatorname{lcm}(q,n)=qn$ and it follows that $q$ and $n$ are coprime.  On the other hand, if $\gcd(q,n)=1$, then $\operatorname{lcm}(q,n)=qn$, we have $g^{nq}=1$ and hence all of  $$ 1,g^q,g^{2q},...,g^{(n-1)q} $$  are distinct again and $g^q$ is a generator.  Is my proof correct at all and is there maybe a more elegant argument?
85287
Suppose $g\in C^1 [a,b]$. Prove that for all $\epsilon &gt; 0$, there is $\delta &gt; 0$ such that $|{g'(c)} - {{g(d)-g(c)} \over {d-c}} |{&lt; \epsilon }$ for all points $c,d \in [a,b]$ with $0 &lt;|d-c|&lt; \delta$   First, I don't know what $C^1 [a,b]$ means.  Some ideas:  By Mean value theorem, $g'(c) =  {{g(b)-g(a)} \over {b-a}} $ since $c\in[a,b]$.   To show $|{g'(c)} - {{g(d)-g(c)} \over {d-c}} |{&lt; \epsilon }$ whenever $0 &lt;|d-c|&lt; \delta$ for $ c,d \in [a,b]$. I guess I have to use the definition of limit of continuous function. But, I don't know how to connect all of these ideas.