Intuition behind the first isomorphism theorem

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

Show that mapping θ(x):a → log a is an isomorphism from R* under multiplication to R under addition.

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

group isomorphism - Do the octonions contain infinitely many copies of the quaternions?

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

group isomorphism - Determine whether there is an onto homomorphism from $(\mathbb{Z}_6,+)$ to $(\mathrm{Z}_3,+)$

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

Confusion about 1st isomorphism theorem

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

Is a factor group an isomorphism of the group it's in?

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

Let $G$ and $H$ be groups. Prove that $G\times H$ is isomorphic to $H\times G$

"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

Banach Algebra Isomorphism

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

Understanding isomorphisms.

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

Word for structurally equivalent group homomorphisms

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

Prove Isomorphism of two groups

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