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5

Hint: $\mathbb{Q}^\times$ has an element of order $2$. Side note: $\mathbb{Q}$ and $\mathbb{Q}^\times$ actually have the same cardinality. One defining feature of infinite sets is that an infinite set and a proper subset of it can have equal cardinalities, so we must be careful in that regard.


3

The group algebra is a functor $\mathsf{Grp} \to \mathsf{Alg}_K$ which is left adjoint to the "group of units" functor $\mathsf{Alg}_K \to \mathsf{Grp}$, $A \mapsto A^\times$. It is defined for all unital $K$-algebras. No non-trivial ring has the property that its underlying semigroup is a group, because $0$ is not invertible. If $V$ is some $K$-module, ...


2

The question is answered by Zhen Lin in the comment. Morphisms of subobjects (resp. quotients) are defined as morphisms of the underlying objects which are compatible with the inclusion morphism (resp. projection morphism). And of course Mac Lane mentions this.


2

Yes this holds for Fermat primes. You need the following observation. This is sometimes called the $N/C$ Theorem. Lemma Let $G$ be a group with a subgroup $H$, then $N_G(H)/C_G(H)$ embeds homomorphically into Aut$(H)$. Here $N_G(H)=\{g \in G : gH=Hg\}$, the normalizer of $H$ in $G$ and $C_G(H)=\{g \in G : gh=hg \text { for all } h\in H\}$, the centralizer ...


1

Let $N$ be a cyclic group of prime order, and let $A$ be any group of automorphisms of $N$. Then the orbits of $N\setminus\{e\}$ under $A$ all have the same size. (The proof is easy, based on the fact that every element of $N\setminus\{e\}$ is a generator of $N$, and hence every element of $N\setminus\{e\}$ is a power of every other element.) Note that this ...


1

$r=3$ for dihedral groups of order divisible by $4$. The standard presentation shows that $r \le 3$, and the fact that the Schur multiplier is nontrivial (it has order $2$) shows that $r-d \ge 1$, so $r=3$.


1

Lagrange's theorem tells us if f $G$ is a finite group and $H$ is a subgroup of $G$, then $\vert H \vert$ divides $\vert G \vert$. As a corollary we get that given $a \in G$, $\vert a \vert$ divides $\vert G \vert$ which means that $\vert G \vert = r \vert a \vert$ for some $r \in \mathbb Z$. As a corollary to that we get that $a^{\vert G \vert} = e$ since ...



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