Partition of a group by proper subgroups with trivial intersection where no two subgroups are isomorphic to each other? I was wondering if there exists a group $G$ and a partition of $G$ by proper subgroups $H_1,\cdots,H_n$, meaning $G = \cup_{i=1}^{n} H_i$ and $H_i \cap H_j = \{1_G\}$ for every $i,j$, such that $H_i \not\cong H_j$ for $i\neq j$. I have already tried to check groups of small order and to use the sylow theorems with no success.
 A: Too long for a comment...
A group partition of a group $G$ is a family of subgroups $\{H_i\}_{i\in I}$ such that $G=\cup_{i\in I}H_i$, and $H_i\cap H_j=\{1\}$ if $i\neq j$.  You are asking for groups with partitions in which any two parts are non-isomorphic.
The study of group partitions was begun by Miller in 1906, and a complete classification of finite groups with partitions was completed by Baer, Kegel, and Suzuki in 1961:

Theorem (Baer, Kegel, Suzuki) A finite group $G$ admits a nontrivial partition if and only if it is one of the following:

*

*$S_4$.

*A $p$-group with $H_p(G)\neq G$ and $|G|\gt p$, where $H_p(G)=\langle x\in G\mid x^p\neq 1\rangle$.

*A group of Hughes-Thompson type.

*A Frobenius group.

*$\mathrm{PSL}(2,p^n)$, where $p^n\geq 4$.

*$\mathrm{PGL}(2,p^n)$, where $p^n\geq 5$ and $p$ is odd.

*$\mathrm{Sz}(2^{2n+1})$.


Above is taken from the introduction to the Atanasov and Foguel paper listed below.
I don't know if any of the above groups would admit a partition like you are asking (we can certainly take a look to see how these groups are partitioned in the relevant papers, if they are), but that restricts the type of groups you need to look at.
References:

*

*Atanasov, R. and Foguel, T. Loops that are partitioned by groups. J. Group Theory 17 (2014), 851-861.

*G. A.  Miller,  Groups in which all the operators are contained in a  series of subgroups such that any two have only identity in common. Bull. Amer. Math. Soc. 17 (1906/1907), 446–449.

*Baer, R. Partitioner endlicher Gruppen. Math. Z. 75 (1960/1961), 333-372.

*Kegel, O.H. Nicht-einfache Partitionen endlicher Gruppen. Arch. Math. 12 (1961), 170-175.

*Suzuki, M. On a finite group with a partition. Arch. Math. 12 (1961), 241-274.

A: Apparently infinite unions are also of interest. Let $G$ be a free abelian group of countable rank, with basis $x_1,x_2,\dots$. Let $Y$ denote the set of all elements $\prod x_i^{a_i}\in G$ where all but finitely many of the $a_i$ are $0$, and the GCD of the non-zero $a_i$ is $1$. Work up to sign, so that the first non-zero $a_i$ is positive. The set $Y$ is the set of all generators of cyclic subgroups that are not proper subgroups of other cyclic subgroups.
Since $Y$ is countable, one may place an ordering $<$ on $Y$, or equivalently order them $y_1,y_2,\dots$. Since any subgroup of $G$ is free abelian, the subgroup generated by $n$ linearly independent elements of $Y$ is free abelian of rank $n$. Let $D_i\subseteq Y$, and $H_i=\langle D_i\rangle$ be defined inductively as follows:

*

*Let $D_1=\{y_1\}$.

*Let $D_i$ have size $i$, and consist of the smallest $y_j$ not contained in the union of the $H_k$ for $k<i$, plus some larger elements of $Y$, chosen so that the union of all $D_k$ for $k\leq i$ is linearly independent.

The union of the $H_i$ must equal $G$, and by construction all intersections are trivial. Since $H_i$ has rank $i$ they are all pairwise non-isomorphic.
A: Let me add some background. There is a lot of research on groups covered by a finite set of proper subgroups. It all started with a seminal paper of Bernhard Neumann, Groups covered by finitely many cosets, Publ. Math. Debrecen 3 (1954), 227-242, in which he proves that if a group is the union of a finite number of subgroups (or even cosets), you only have to take into account those subgroups having finite index. This implies that a group has a finite covering by subgroups if and only if it has a finite non-cyclic homomorphic image. See also work of Mira Bhargava, the mother of the brilliant number theorist Manjul Bhargava.
A: There are no Abelian groups with this property. This follows from
the following statement.

If $G$ is a finite abelian group and $G=\cup_i H_i$, where all $H_i$ are
proper subgroups of $G$ and $H_i\cap H_j=\{1\}$ for $i\neq j$, then
$G$ is an elementary abelian $p$-group and $|G|>p$.

Proof. Let $x\in G$ be an element of maximal order and $o(x)=m$ ($o(x)$ is the order of $x$).
Assume that $m$ is not prime and bring this assumption into contradiction.

*

*If $y\in G$, then $o(y)$ is a divisor of $m$.


*Let $x\in H_1$. Let $y\in H_2$, $y\neq1$, such that $s=o(y)<m$. (If there is no such $y$, then $m$ is prime.)


*We have $y\notin H_1$. Therefore $xy\notin H_1$. Let $xy\in H_i$, $i>1$. Since
$$
(xy)^s=x^s
$$
and $H_1\cap H_i=\{1\}$ it follows that $x^s=1$. Contradiction.
Addition.
Since the question arose about infinite groups, I would like to point out that our statement is also true for them.

If $G$ is an infinite abelian group and $G=\cup_{i=1}^n H_i$ and
$H_i\cap H_j=\{1\}$ for $i\neq j$, then $G$ is an elementary abelian
$p$-group.

The following three cases are possible: 1) the group $G$ has elements of infinite order and they do not lie in one $H_i$; 2) all elements of infinite order lie in one $H_i$ and then there are elements of finite order; 3) all elements of $G$ have finite order.

*

*Suppose that $x,y$ are elements of infinite order and
$x\in H_1$ and $y\in H_2$. Consider elements of the form $xy^k$, $k\in\mathbb{Z}$. There exist such $k$ and $l$, $k\neq l$, that $xy^k,xy^l\in H_i$ for some $i\geq3$.
But then $y^{k-l}\in H_i\cap H_2$. Contradiction.

Note that this is the only place where it is used that the number of subgroups $H_i$ is finite.


*Let all elements of infinite order lie in $H_1$ and let $x\in H_1$ be an element of infinite order. Let $y\in H_2$ and $y^m=1$, $m>0$.  The element $xy$ has infinite order and $xy\notin H_1$. We have a contradiction with point 1.


*If all elements of $G$ have finite order, then by virtue of the finite case every finite subgroup of $G$ is an elementary abelian $p$-group for some simple $p$. Then $G$ is also elementary abelian.
