I've been studying homomorphisms and have got to the first isomorphism theorem. The proof generally makes sense but I'm not able to see why the theorem is intuitively true, nor how it could be useful.I was hoping someone could go through the proof and explain the 'intuitiveness' of each step and why it should 'feel' true. So the theorem and the structure of the proof I've is as follows:Let $\phi: G_1 \rightarrow G_2$ be a group homomorphismLet $\ker(\phi)$ be the set $\left\{x \in G_1 | \phi(x) = 0_{G_2}\right\}$ Then $\left\{ y \in G_2 | \exi...Read more

An exercise in a book states that "Show that mapping θ(x):a → log a is an isomorphism from R* under multiplication to R under addition."What should be done first?Edit (interpreted question): Show that the mapping $\theta : (\mathbb{R}_{>0}, \cdot) \to (\mathbb{R}, +)$, defined by $\theta(x) = \log(x)$ for all $x \in \mathbb{R}_{>0}$, is an isomorphism....Read more

Suppose you have a group presentation $G=<a,b|a^{5}=b^{2}=e, ba=a^{2}b>$. In general, how do you find the group it is isomorphic to?I've seen examples using the Fundamental Theorem of Finitely Generated Abelian Groups but we haven't even covered this in class yet. Is there another procedure for finding the group to which this presentation is isomorphic?...Read more

I have created the two tables but can not find a one to one correlation between the values in the two tables. I would appreciate it if anyone can point me in the right direction to understand how to solve this. +4 0 1 2 3 0 0 1 2 3 1 1 2 3 4 2 2 3 0 1 3 3 4 1 2 *5 1 2 3 4 1 1 2 3 4 2 2 4 1 3 3 3 1 4 2 4 4 3 2 1Thank you!question 31.3-1...Read more

Note: by "infinitely many", I'm confident I always mean $\beth_1$ many herein.We can easily show the quaternions contain infinitely many copies of $\Bbb C$ because, given any unit vector $\in\Bbb R^3$ of components $b,\,c,\,d$, $\Bbb R[h]$ is isomorphic to $\Bbb C$ with $h:=bi+cj+dk$. Sure, these aren't "independent" copies of $\Bbb C$ in the same way $\Bbb R[i],\,\Bbb R[j],\,\Bbb R[k]$ are. But it's still of interest because, for example, a family of tensor products over matrices using different copies of $\Bbb C$ provide an easy definition of...Read more

This question already has an answer here: Cayley tables for two non-isomorphic groups of order 4. 2 answers...Read more

I have proven that every discrete subgroup of $\mathbb{R}^n$ is isomorphic to a $\mathbb{Z}^m$, but with no condition on $m$ (potentially, $m>n$). How do I prove that $m\leq n$?I have tried doing it through an induction, which seems easiest but I can't really get the induction down (basis is easy)....Read more

Question: We have to determine if there exists a homomorphism from $(\mathbb{Z}_6,+)$ onto $(\mathrm{Z}_3,+)$. My efforts: Let $\phi$ be an onto homomorphism. Since $\phi$ is surjective, then by the first isomorphism theorem, $\mathbb{Z}_6/\ker\phi \cong \mathrm{Im}(\phi)=\mathbb{Z}_3$. What can I say after this?Added: Can we say? $\mathbb{Z}_6/\ker\phi \cong \mathrm{Im}(\phi)=\mathbb{Z}_3\implies \left|\mathbb{Z}_6\right|=|\ker\phi||\mathbb{Z}_3|$. Contrapositively, $|\ker\phi||\mathbb{Z}_3|\neq\left|\mathbb{Z}_6\right|\implies $ $\phi$ ...Read more

I was thinking about the first isomorphism theorem, and something curious occurred to me. If φ is an injective homomorphism from group G to group H, then the kernel is trivial. But then G/kerφ is isomorphic to H, yes? But if kerφ = {e}, then isn't G/kerφ = G/{e} = G? But then wouldn't that make G isomorphic to H?That doesn't make sense though because that would imply that every injective homomorphism is an isomorphism. What am I doing wrong?...Read more

The definition for factor groups that I was given in my textbook (A First Course In Abstract Algebra by John B. Fraleigh) was: If $N$ is a normal subgroup of a group $G$, the group of cosets of $N$ under the induced operation is the factor group of $G$ modulo $N$, and is denoted $G/N$. The cosets are residue classes of $G$ modulo $N$.After reading the properties section from the wikipedia article on factor groups, I know that the $G/\{e\}$ is isomorphic to $G$, but I am wondering why other factor groups are not isomorphic to $G$? Since the fac...Read more

"A Book of Abstract Algebra" by Dr. Pinter presents this exercise: Let $G$ and $H$ be groups. Prove that $G\times H$ is isomorphic to $H\times G$. My understanding is that I need to find a function $f$, such that $f(G\times H) = H\times G$.My initial thought was to rely on the commutative property of multiplcation. In other words, the function $f$ would be the identity function.But that does not seem right. Please point me in the right direction.Note - I looked at related questions, but did not recognize them (perhaps due to terms that I don't...Read more

Let $X=l^1$ with coordinate multiplication - it is commutative Banach algebra without unit. The Gelfand transformation is defined as $\widehat{x}(e_n)=x_n$ for $x \in l^1$. I would like to prove, that this algebra is isomorphic to $C^*$-algebra. Any ideas ? Thanks....Read more

Can someone explain this solution to me?The goal is to find an isomorphism from $ [\mathbb{Z}_6, \oplus_6]$ to $[\mathbb{Z}_7\setminus\{0\}, \otimes_7]$, where ${}\setminus\{0\}$ means "excluding 0".There are 2 isomorphisms: $f(0) = 1, \quad f(1)=3, \quad f(2)=2, \quad f(3)=6, \quad f(4)=4, \quad f(5)=5$and $f(0) = 1, \quad f(1)=5, \quad f(2)=4, \quad f(3)=6, \quad f(4)=2, \quad f(5)=3$I don't understand part of the solution...I know the generators for $\mathbb{Z}_6$ are {$1$, $5$}. Similarly, the generators for $\mathbb{Z}_7\backslash\{0\}$ ...Read more

I am to determine some groups and maps in a diagram. I obviously only have to determine the groups up to "structural equivalence" i.e. isomorphism. Is there an equivalent word for structural equivalence in homomorphisms? For example, the $\mathbb{Z} \rightarrow \mathbb{Z}_6$ homomorphisms $x \mapsto x \mod 6$ and $x \mapsto -(x \mod 6)$, mapping $1,2,3,4,5$ to $1,2,3,4,5$ and $5,4,3,2,1$, respectively, are equivalent, in the sense that $1$ is equivalent to $5$, $2$ to $4$ and so on, and these elements are equivalent in the sense that there is a...Read more

Suppose that $G_1$ and $G_2$ are finite groups and $\beta: G_1\to G_2$ is an isomorphism. If $x_2 = \beta(x_1)$ for a given element $x_1 \in G_1$, prove that $x_1$ and $x_2$ have the same order.I think I should use induction but I am not sure how could I prove it?...Read more