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8

Let $G=S_3$ be the group of permutations on three elements, and $H$ be the subgroup of even permutations. This answers yes to both questions.


6

Sure, here's an answer to both. Consider the quaternions, $Q_{8}$, and the normal subgroup $H = \langle i \rangle$. Then $Q_{8}/H \cong \mathbb{Z}_{2}$, but $Q_{8}$ is neither cyclic nor abelian. Here's another more general example. Consider $D_{2n} = \langle r, s \mid r^{n}=s^{2} =1, rs=sr^{-1}\rangle$. The subgroup $H = \langle r \rangle$ is always ...


5

If two subgroups of a group are normal, have trivial intersection, and generate the group via elements of the form $xy$ with $x$ in the first group and $y$ in the second, then the group is the direct product of the two groups. The group of order $q^2$, is abelian, so the result follows.


4

There's a little mistake $x*(y*z) = x *(y+z -1) = x+((y+z-1)\color{red}{-1}) = x+y+z-2$


4

Hint: Since $G$ is abelian, $x \mapsto x^{-1}$ is an automorphism.


4

If $G$ is abelian, consider the map $x \to x^{-1}$.


4

Here is a proof that doesn't involve going through $\mathbb{Q}$ (and works for any PID): The image of $f$ is a submodule of a free module, so it is itself free (since $\mathbb{Z}$ is a PID). Therefore the short exact sequence $0 \to \operatorname{ker} f \to G \to \operatorname{im} f \to 0$ is split, and therefore (general fact about split exact sequences of ...


4

Am implication of this is that every finite abelian group is isomorphic to some cyclic group, which is not true.


3

$\textbf{HINT-}$ Consider $\mathbb{Z_4}$ and $\mathbb{Z_2 \times Z_2}$. So FALSE.


3

Not always. Counter example: $$G=\Bbb Z_2\times \Bbb Z_2\;,\;\;\;H=\langle\;(1,1)\;\rangle$$


3

The big group has, as you pointed out, only $p^2$ elements!


3

Yes, for morphisms between free Abelian groups this holds. Just interpret the matrix with respect to some basis as one with rational coefficients, and use its rank, and kernel dimension. Formally this means applying the functor of tensoring with the rational numbers. Note however that much interesting things about Abelian groups involves torsion, which is ...


3

You have to calculate how many groups elements are represented by words at length at most $N$ in the group generators. Let $G$ be free abelian with generators $a,b$. How many elements have shortest representative of length $n$? For $n=0$ there is just $1$. For $n>0$, we have $a^n,b^n,a^{-n},b^{-n}$, which is $4$. We also have $a^kb^{n-k}$ for ...


3

Example: the quaternion group $Q$ of order $8$ and the dihedral group $D_4$ of order $8$ have the same conjugacy class structure $(1+1+2+2+2)$, yet these non-abelian groups are not isomorphic.


3

We want to prove by induction on $n$ that every subgroup of an abelian group generated by $n$ elements is finitely-generated. If $n=1$, it is clearly true. From now on, suppose it is true for $n$. Let $G= \langle a_1, \dots, a_{n+1} \rangle$ be an abelian group, $H \leq G$ be a subgroup and $\rho : G \to G/ \langle a_{n+1} \rangle$ be the quotient map. ...


3

Without (directly) using the structure theorem for f.g. abelian groups, you can proceed as follows. Suppose the result holds for all $m'| m$, and choose some prime $p$ dividing $n > 1$. By Cauchy's theorem, there exists an element $\gamma\in G$ of order $p$. Put $N = \langle{\gamma}\rangle\subset G$ and $Q = G/N$. Then $Q$ has order $m/p$ and thus ...


3

Your question/answer and what you write at the bottom contradict one another. There are $3$ abelian groups of order $8$, up to isomorphism. And you've listed them: $$\mathbb Z_8,\;\;\mathbb Z_4\times \mathbb Z_2,\, \text{ or }\,\mathbb Z_2\times \mathbb Z_2\times \mathbb Z_2.$$ Among those three groups, no two of them is isomorphic. So it is false any ...


2

Given an abelian group $G$ its set of endomorphisms $$ {\rm End}(G)=\{f:G\rightarrow G,\text{homomorphism}\} $$ is a ring under the usual operations of sum and composition, with unity the identity map. Given any ring with unity $R$ there is a unique map of unitary rings $$ \Bbb Z\longrightarrow R $$ given by $n\mapsto n\cdot1_R$ with the usual convention ...


2

The conditions $\sigma\circ\psi_{G}=\varphi_{G}$ and $\sigma\circ\psi_{H}=\varphi_{H}$ are determining for $\sigma$. This because: ...


2

Your instructor's intuition is correct, but we can formalize this by writing down an explicit isomorphism $\tilde{\Phi}: \mathbb{Z}^3 / \langle(1, 1, 1)\rangle \to \mathbb{Z}^2$. First, given an element $(a, b, c) \in \mathbb{Z}^3$, it is equivalent to exactly one element of the form $(x, y, 0)$, namely $$(a, b, c) + (-c) \cdot (1, 1, 1) = (a - c, b - c, ...


2

Hint: For any two elements $g,h$, there's $\phi,\theta: \Bbb Z\to G$ such that $$\phi(1)=g,\theta(1)=h.$$ What does it mean that $\phi\cdot\theta(2)=\phi(2)\cdot\theta(2)$?


2

Proceed in three steps. Show the result for cyclic groups. Show that if $m \mid n_1 \dots n_r$ then $m=m_1 \dots m_r$ with $m_i \mid n_i$ for each $i$. Use the structure theorem for finite abelian groups to tie things together.


2

Let $a \in N$, $b,c \in G$. Since $\langle a \rangle$ is normal in $G$, it is normalized by $b$ and $c$. The automorphism group of a cyclic group is abelian, so $b^{-1}c^{-1}bc$ centralizes $\langle a \rangle$. Now this is true for all $a \in N$, so $b^{-1}c^{-1}bc \in C_G(N) \le N$.


2

You have $G\cong G^2=G\times G\cong G\times G^2=G^3$ (abelian or not, in any category with finite products)


2

I would suggest working in stages. Assuming we already know all about integer arithmetic, first define multiplication for naturals only, with the recursion equations $$ f(0_{\mathbb N},a) = 0_G \qquad f(n+1,a) = f(n+a)+_G a $$ and now prove that it associates and distributes as we want to, as long as the numbers involved are nonnegative. Your question ...


2

Linear algebra over ℚ actually represents an ancient idea of commensurability. Although tensor products are not linear algebra, this way of thinking helps enough for understanding “⊗ℚ”, with some application of order theory. First of all, how should we visualize elementary tensors? Consider for clarity only positive ones, i.e. representable as ℓ ⊗ ℎ where ...


1

Because if the group is not abelian, it does not necessarily have any non-trivial characters at all. For instance, if the group is simple, i.e. it does not have any normal subgroups (for instance, the alternating group $A_5$), then it only has the trivial character, and then of course you cannot have $\chi_0(x)\neq 1$. A non-abelian simple group $G$ only ...


1

Yes, these statements are equivalent, and you're approach is right. Note only that in order to describe an element of $C/C^{\prime}$, when the latter is realized by $C^{\prime}$-cosets in $C$, you have to put $h(a + A^{\prime}) := f(a) + C^{\prime}$. However, more conceptually you should try to prove the equivalence using universal properties of kernel and ...


1

For (2) $\Rightarrow$ (1): Take $a'\in A'$, we want to define an element in $C'$. The only map available to use is $i:A'\to A$ which gets you to $A$. As I mentioned in the comment, one way to get something in $C'$ is to get $[0]=0+C'\in C/C'$. So we push our element $i(a')\in A$ to element in $C/C'$ there is only one way to do it: so you have ...


1

Hint. Your group is abelian, and there are three abelian groups of order $8$: $C_2\times C_2\times C_2$, $C_4\times C_2$ and $C_8$, so it must be among them. These are easily distinguished by looking at the orders of elements (as in the example you showed). For instance, if your group has an element of order $8$, then it must be the cyclic group $C_8$. ...



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