Let $G$ be any group. Let $H$ consist of the same set of elements as $G$, but with a new operation given by $a ∗ b = ba$, for all $a$ and $b$. Show that $H$ is a group, and that it is isomorphic to $G$.  I am having trouble proving what is being asked of me. First, I have no information about how the operation is in $G$. I suppose I could define the operation in $G$ as is usually done for group definition, this is if $a, b \in G$, then its product is $ab$. Based on that, $H$ with the new operation would be a group as well, since they have the same elements. The truth seems like nonsense to me but I can't think of anything else. Also I have tried to establish an isomorphism and let the rest follow from this fact, but so far it has been unsuccessful. Any hint would be helpful.
122411
Let $G = (G,\times)$ be a group, and $G^{op} = (G,o), a \ o \ b = b \times a$ be it's opposite group.  I need to prove that there is a group isomorphism $f: G \to G^{op}$ Obviously, $id_G: G \to G$ is a bijection, but it's a group homomorphism if and only if $G$ is commutative.
77790
Theorem: If $G$ is a finitely generated group, then it has a finite (maybe zero) number of subgroups of index $n$ for any $n\in \mathbb{N}$.   is a sketch of a proof. It consists of assuming such a subgroup $H$ exists, and allowing $G$ to act on $G/H$ by left multiplication. Such action defines a homomorphism $$\phi :G \rightarrow \text{Sym}(G/H)\cong \mathbb{S}_n$$  If $G=\langle g_1, \ldots ,g_r\rangle$, then the homomorphism is uniquely determined by where it sends the generators $g_i$. In particular, there are at most $(n!)^r$ such homomorphisms.  The last step of the proof requires one to notice that a different subgroup $H'$ of index $n$ in $G$ would produce a different homomorphism, and here is where I'm struggling. I've been able to show that $\text{ker}(\phi )=\text{core}(H)$, yet haven't been able to rule out the possibility of $\text{core}(H)=\text{core}(H')$.
82071
If $f:G \rightarrow H$ is group homomorphism, and $H$ is abelian, then $G$ is abelian.  Is that statement correct? Here's my attempt of proof:  Let $a,b \in G$, then: $ab=f^{-1}f(ab)=f^{-1}f(a)f(b)=f^{-1}f(b)f(a)=f^{-1}f(ba)=ba$  Is this correct? I relied on first on $f$ being group homomorphism, then $H$ being abelian.
23462
Let $S$ be the collection of all groups. Deﬁne a relation on $S$ by $G \sim H$ iﬀ $G ≈ H$. Prove that this is an equivalence relation. So $S$ is partitioned into isomorphism classes.   Proof: Let $S$ be a relation on S defined by $G \sim H$ iﬀ $G ≈ H$. Show reflexive, symmetric, transitive. It is obvious that $G≈G$ so $G\sim G$.  Assume $G\sim H$, thus $G≈H$, which implies that $H≈G$, so $H\sim G$, thus symmetric.  Assume $G\sim H$ and $H\sim K$, for $K$ a group in $S$. Thus $G≈H$ and $H≈K$. (I have already shown that if $G≈H$ and $H≈K$, then $G≈K$). From earlier this implies that $G≈K$, so $\sim$ is an equivalence relation.  Thus $S$ is partitioned into isomorphism classes. QED.
34853
Consider the abelian groups $G_n=(\mathbb R^n,+)$ for $n\geq1$.     Claim: For any $n$ and $m$ the groups $G_n$ and $G_m$ are isomorphic.   This claim is true if one assumes the axiom of choice, and I have sketched a proof below. But this claim seems much weaker than the axiom of choice, which brings me to the following questions:   Are there any milder extensions of ZF that make the claim true? If the claim can be proven with some other axioms, what does the proof look like? Are there any known  axioms that make the claim false? Does the claim have any interesting (set-theoretical) corollaries?   This claim was also considered in  earlier post, but from a different point of view.    A sketchy proof of the claim:  $\newcommand{\Q}{\mathbb Q}$ The groups $G_n$ are naturally vector spaces over $\Q$. It is easy to check that the claim is equivalent with $G_n$ and $G_m$ being isomorphic as vector spaces over $\Q$.  All vector spaces have a Hamel basis; let the basis of $G_n$ be $B_n$. Thus $G_n=\Q^{(B_n)}$ (brackets meaning only finitely many nonzero components). But $|\Q^{(B_n)}|=|B_n|$ since the basis is infinite, so $|B_n|=|G_n|=|\mathbb R|$.  There is a bijection between the bases of the vector spaces $G_n$ and $G_m$ since they have the same cardinality. This gives a linear bijection between them.
34805
Let $G$ be a group and $a,b\in G$.  Show $(ab)^n=a^nb^n$ for all $n\in\mathbb{Z}$ if and only if $ab=ba$.  I don't known where to start. It seems trivially.
1204
I am trying to solve this problem:  Let $G$ be a finite and simple group, and let $p$ be a prime number such that $p$ divides $|G|$. If $n_p(G)=n$ for $n&gt;1$ (n_p denotes the number of p-Sylows) then $G$ is isomorphic to a subgroup of $\mathbb A_n$.  What I thought of was the following:  I consider the action of $G$ on $X$={H subgroup of G, H p-Sylow} by conjugation. Then there is a morphism $$\psi:G \to S(X)$$ $$g \to gSg^{-1} \space \forall S \in X,$$  where $S(X)$ is the set of all bijective functions $f:X \to X$. First of all, I would like to affirm $S(X) \cong \mathbb S_n$, but I don't know if this is true. Second, note that  $Ker(\psi) \neq G$ because if not $n_p=1$ and $Ker(\psi)$ can't be a proper subgroup because if that was the case $Ker(\psi) \lhd G$, then we have $Ker(\psi)=\{e\}$.  By the first isomorphism theorem, $G \cong G/Ker(\psi) \cong Im(\psi)$. Now, if I could show that $Im(\psi)$ is isomorphic to a subgroup of $\mathbb A_n$, then I would be done.  I would appreciate any hints or suggestions.
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 $(aH)(bH)=abH$."
39622
Prove that any two cyclic groups of the same order are isomorphic.   Let the groups be $G,H$ with order $k$. Let $G=&lt;a&gt;$ and $H=&lt;b&gt;$. Thus we have $|a|=|b|=k$ and by definition, $G=\{a^0,a^1,...,a^{k-1}\}$ and $H=\{b^0,b^1,...,b^{k-1}\}$. Next I set up the bijection $\theta : G \rightarrow H: \theta(a^n)=b^n$. I think I should proceed by showing $\theta$ is an isomorphism but I'm not sure how to do that. What's a good way to go about showing $\theta$ is an isomorphism?
69340
Question:  Give an example of a group $G$ having a subset $H$ such that $HH=H$, where $HH = \left\{h_1h_2\mid h_1,h_2\in H \right\}$, but $H$ is not a subgroup of $G$.   Attempted Solution:  I haven't gotten far with this problem. In a previous exercise, I was able to prove that $HH=H$ if $H$ is a subgroup of $G$. I am having a hard time thinking of a scenario where $h_1h_2\in H$ and $H$ is not a group. I've played around with making $h_1h_2$ not closed under its operation, but I haven't gotten anywhere yet.
79458
Following the advice given in , I have started to study Group Theory from the very basics. My reference text is Abstract Algebra by Dummit and Foote. While going through the exercises (page 24) I found one problem which required more effort than others:  Assume $G = \{1, a, b, c\}$ is a group of order $4$ with identity $1$. Also assume that $G$ has no elements of order $4$. Use the cancellation laws to show that there is a unique group table for $G$. Deduce that $G$ is abelian.  I proceeded with solution as follows. Since $G$ is of even order there must be at least one element of order $2$ (this statement itself was one of the problems in the exercises and is easy to handle). From the wording of the question it seems that none of the elements $a, b, c$ is singled out with specific properties and hence $a, b, c$ must behave in exactly the same manner. Hence each of them is of order $2$. Thus $a^{2} = b^{2} = c^{2} = 1$.  Next I analyze product $ab$. Clearly it can't be $a$ (as $b$ is non-identity), or $b$ (as $a$ is non-identity) or $1$ (because $ab = 1 = aa$ implies $b = a$). Hence $ab = c$ and similarly $ba = c$. By the similar nature of all elements $a, b, c$ it follows that $bc = cb = a$ and $ca = ac = b$. Thus the operation of the group is defined properly for all elements and clearly it is abelian.  I think my solution is OK, but I am not sure. Note that I have not used the fact that $G$ has no elements of order $4$. Please let me know :  1) if my solution is correct. If not then point out the flaws.  2) if there is any better / shorter solution. If so provide hints and not solution.  The question may sound too easy / simple but I request to treat me like a beginner who has never heard of Group Theory and is reading fresh from the reference text I mentioned. Thanks in advance for your inputs.  EDIT: Thinking further about this problem I wondered what would happen if an element of order $4$ was allowed in $G$. In that case I think that the group has to be isomorphic to the group $G_{1} = \{1, 2, 3, 4\}$ with modulo $5$ multiplication as the group operation. The argument is as follows. Suppose $a$ is of order $4$ and let $b$ be its inverse then $b$ is also of order $4$. Now $c$ has to be its own inverse so that $c^{2} = 1$. Again $a^{2} \neq 1$ (as $ab = 1$), $a^{2} \neq a$ (as $a \neq 1$), $a^{2} \neq b$ (as it would mean $b^{2} = 1$ and thus $b$ would be of order $2$). Hence $a^{2} = c$ and similarly $b^{2} = c$. Next we can see that $ac \neq a$, $ac \neq c$, $ac \neq 1$ (as $ab = 1$) so that $ac = b$. Similarly $ca = b$, $bc = a$, $cb = a$. Also it is clear that this group turns out to be cyclic with both $a$ as well as $b$ as generators. And it also follows that there are the only two ($G$ in original question and $G$ in new variation) groups of order $4$ upto isomorphism.  I hope I have learnt something from the answers given for the original question and this solution is correct. Please let me know if there is any problem with this reasoning.
72149
Let $G$ and $H$ be groups and assume that there exist onto homomorphisms $$\phi :G\rightarrow H,\qquad\psi :H\rightarrow G.$$ If $H$ and $G$ are finite then it is clear that $H$ and $G$ are isomorphic. It is also valid for finitly generated abelian groups. However, in general this is not true even in the abelian case.     My question is: if we know that $G,H$ are finitly generated and such $\phi ,\psi$ exist then are $G$ and $H$ isomorphic?
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!
15770
The following could have shown up as an exercise in a basic Abstract Algebra text, and if anyone can give me a reference, I will be most grateful.  Consider a set $X$ with an associative law of composition, not known to have an identity or inverses. Suppose that for every $g\in X$, there is a unique $x\in X$ with $gxg=g$. Show that $X$ is a group.  Note that I’m not asking for a proof (though a really short one would please me!), just some place where this has been published.
5675
I have the following question:      Let $G$ be a group and let $H$ be a subgroup of finite index of $G$. Let $|G:H|=n$   Then it holds: $g^{n!}\in H$ for all $g\in G$.   Why is this true?  I think, that's not very difficult, but I have no idea at the moment.   Thanks!
56220
Let $G$ be a group and let $c$ be a fixed elements of $G$.    Now, I'm going to define a new operation "*" on $G$ by $a*b=ac^{-1}b$    How do I prove that the set $G$ is a group under *.  Thanks for the help!
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