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7

The obvious homomorphism $\phi: G\to G/H\times G/K$ is injective, since the kernel is the set of elements in $H\cap K$. Since both quotients are abelian, the product is abelian, and $G$ is isomorphic to a subgroup of the product, and is thus abelian itself.


3

Let $a, b \in G$. Consider $[a, b]:=aba^{-1}b^{-1}$ which is frequently called the commutator. (Up to choice of convention.) Then the hypotheses tell us that $[a, b] \in H$ and $[a, b] \in K$ because $ab=ba$ mod $H$ or $K$.


3

For $p$ groups, $G'\leq \Phi(G)$. Thus, $HG'=G$ then $\Phi(G)H=G\implies H=G$. Note: $\Phi(G)$ is the Frattini subgroup of $G$ and it is the intersection of all maximal subgroups of $G$. Secon Way: If you are not familiar with Frattini Subgroup, You can follow this elemantary proof. Let $M$ be a maximal subgroup of $G$ then $|G:M|=p$ and $M$ is normal ...


2

$\mathbb{Z}_{p}^{m}$ is a vector space of dimension $m$ over $\mathbb{Z}_{p}$. Any single generator generates a 1-dimensional subspace. If you want to span the entire vector space, you certainly need $m$ linearly independent vectors.


2

If $G$ is a group of order $pq$ where $p>q$ are prime and $q$ doesn't divide $p-1$ then $G$ is cyclic.(application of Sylow's theorem).


2

There are at most $\;\aleph_0\cdot\aleph_0=\aleph_0\;$ finite words that can be formed out of the elements of each group $\;G_\alpha\;$ . Every word in that free product is the concatenation of a finite number of words ( namely, from a finite subset of $\;\{G_\alpha\}$ ) . Thus, we have at most $$\aleph_0\cdot\aleph_0=\aleph_0\;\;\text {words ...


1

Several examples of finite loops (quasigroups with left and right identity) of small order with some additional properties can be found here. Another source showing several examples of finite loops Cayley tables. A third example of very group-like loop which is not a group, answer to a related Math.SE question. Again, in the absence of associativity, ...


1

The Jacobson radical. Take a noncommutative ring $R$ with 1. Any left ideal is either contained in another or is maximal. The elements common to all maximal left ideals, i.e. $$ J = \bigcap_i M_i, $$ is a group in two ways: It is an abelian group because it is an ideal (inherits group additivity from $R$, pretty obvious). It is group under circle ...



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