Let’s call a group $G$ invertible, if $\forall H \triangleleft G$, there $\exists K \triangleleft G$, such, that $K \cong \frac{G}{H}$ and $H \cong \frac{G}{K}$  All finite abelian groups are known to be invertible. All simple groups also are invertible.  However the vast majority of finite groups seems to be non-invertible: that includes symmetric groups, dihedral groups, holomorphs of cyclic groups, generalised quaternion groups and so on…  So my question is:     Is there some sort of classification of invertible finite groups?
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A common mistake for beginning group theory students is the belief that a quotient of a group $G$ is necessarily isomorphic to a subgroup of $G$.  Is there a characterization of the groups in which this property holds?  If this question is too broad, I might ask if such a characterization exists for $p$-groups.    History: I originally posed the opposite question, regarding groups for which $\exists N\unlhd G\,:\, \not\exists H \unlhd G\, \text{  s.t. } H \cong G/N$, and crossposted this to .  I received an answer there to the (now omitted) peripheral question about probability, which shows that most finite groups probably have this property.  After this, I changed the question to its current state, as this smaller collection of groups is more likely to be characterizable.
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The question  got me curious and thinking about the following:   For which $n$ and $k$ are there nontrivial homomorphisms from the symmetric group $S_n$ to the cyclic group $\mathbb{Z}/k\mathbb{Z}$ of order $k$? I'm thinking it might be an open problem. Is this a problem with a known solution? If so, how can we prove it?   Thanks!
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I just want to make sure that every group $\mathbb{Z}_n$ for any $n$ is cyclic. Further, every group of prime order is cyclic because it is isomorphic to $\mathbb{Z}_p$. I think I have a handle on that, but what other finite groups are there that aren't cyclic? In particular, I'm looking for examples that aren't direct products of groups, just simply groups like $\mathbb{Z}_n$, not $\mathbb{Z}_a \times \mathbb{Z}_b$, etc.
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Let $G$ be a group and $H&lt;G$ such that $[G:H]&lt;\infty$. There exists a subgroup $N\triangleleft G$ such that $[G:N]&lt;\infty$.  I have to show this fact (that according to my book is due to Poincaré), but I think that the statement, written in this way, is trivial: for every group $G$, I can take $N=G$, in fact $G\triangleleft G$ and $[G:G]=1$.  Where am I wrong?  If I'm not wrong, do you know a similar statement?
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I'm looking for an example of $2$ non isomorphic groups $G_1,G_2$  that are finitely generated and presented that have the same finite quotients (up to isomorphism) .  Thanks
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Let $G$ be a solvable group and $f: G \rightarrow H$ an isomorphism. My question is if $H$ is a solvable group.   Let $1=G_0 \trianglelefteq G_1 \trianglelefteq ... \trianglelefteq G_n = G$ be a normal series of $G$ whose factors are abelian groups. Clearly, $1=f(G_0) \trianglelefteq f(G_1) \trianglelefteq ... \trianglelefteq f(G_n) = H$ is a normal series of $H$ but are the quotients abelian?  Is there a weaker condition for invariance?
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So I've been working on this problem for a while. I just can't seem to find the solution. Also in the previous subquestion, I showed that $G$ is abelian iff $[G,G] = {e}$. Where $e$ is the neutral element of $G$ and $[G,G]$ is the commutator group of $G$.  Let $H\triangleleft G$, Show that $G/H$ is abelian iif $[G,G] \subseteq H$.  I've proved this direction  $\Rightarrow $) If $G/H$ is abelian, $\forall x,y \in G$,  $xHyH = xyH = yxH = yHxH$.   Therefore have that $G$ is abelian. As shown earlier, $[G,G] = {e}$ if $G$ is abelian. So $[G,G] = {e} \subseteq H$.  I can't seem to figure out the other direction...  $\Leftarrow$)  I only have that for $x,y \in G$,  $x^{-1}y^{-1}xy \in H$ and that $H\triangleleft G$. It does not seem to lead me anywhere.  Thanks for the help!
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I was trying to find an example such that $G \cong G \times G$, but I am not getting anywhere. Obviously no finite group satisfies it. What is such group?
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If $G$ is a group and $G'$ is generated by $\{xyx^{-1}y^{-1}|x,y\in G\}$, then $G'\trianglelefteq G$ and  $G/G'$ is Abelian.  At first, I thought this is easy because I thought $g[x,y]g^{-1}=[gxg^{-1},gyg^{-1}]$. But it was a mistake. I could not prove that it is also subgroup of G and $G/G'$ is abelian. I need help.
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When people want to classify a group with certain (small) order, they seem to find a normal subgroup $H$ and a subgroup $K$, and then consider how many distinct $H \rtimes_\varphi K$ are there.  My question is: For any group $G$, does it always exist normal subgroup $H$, subgroup $K$ and $\phi : K \rightarrow Aut(H)$ s.t. $G\cong H \rtimes_\varphi K $?  Sorry if this question is too basic...  Thanks the comment and the answers. Then if the $G$ is not simple, can we have can $G\cong H\rtimes_\varphi K$ for proper subgroup $H$, $K$, with $H$ normal?  EDIT:  If we have a group $G$, that is not simple (then $G$ has at least one normal subgroup), are there always a non-trivial normal subgroup $H$ and a proper subgroup $K$ s.t. $G \cong H \rtimes K$. (Note that if $G$ is simple this is clearly not possible).  But the quaternion group and $Z_4$ given in answers have shown that we cannot always have it.
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I will try and show that if our group $G$ is not cyclic then it is isomorphic to .  So suppose $G$ is not cyclic. Then its elements, excluding $e$, are of order $3$ or $2$. Supposing there is an element of order $3$ gives three distinct elements $g, g^2, g^3$. By hypothesis there exists  $h \in G$ such that $h$ is distinct from these three elements. So we can make $gh$ which must be equal to one of the elements previously mentioned. Setting this equal to the various elements:  \begin{align} &amp;gh=g \Rightarrow h=e &amp;&amp; gh=g^2 \Rightarrow h=g &amp; \\ &amp;gh=h \Rightarrow g=e, &amp;&amp; gh=g^3 \Rightarrow h=g^2 &amp; \end{align} Each of which is a contradiction. Therefore all the elements are of order two and we can get an isomorphism between $G$ and $H \times H$ where $ H$ is cyclic of order two. But this direct product is isomorphic to Klein's four-group so $G$ is isomorphic to $V$.  Is this correct?  Also it is just given in the book that $V$ is isomorphic to this product; how do I show this without checking that $f(gh)=f(g)f(h)$ for all combinations?
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Is the following statement true or false:  If $G$ is a group with the property that $g=g^{-1}$ for all $g \in G$, then $G$ is abelian.   I believe it is false since I know that abelian or commutative property implies that every element in $G$ must have an inverse. Thus $g\cdot g^{-1}= g^{1}\cdot g = e$. I need someone to check my attempt.
98600
I was wondering this today, and my algebra professor didn't know the answer.      Are subgroups of finitely generated groups also finitely generated?    I suppose it is necessarily true for finitely generated abelian groups, but is it true in general?  And if not, is there a simple example of a finitely generated group with a non-finitely generated subgroup?  NOTE:  This question has been merged with another question, asked by an undergraduate.  For an example not involving free groups, please see Andreas Caranti's answer, which was the accepted answer on the merged question.
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In , The Chaz asks whether $G\times G\cong H\times H$ implies that $G\cong H$, where $G$ and $H$ are finite abelian groups. The answer is to his question is yes, by the structure theorem for finite abelian groups, as noted in the answer by Anjan Gupta.  Even though I don't know the first thing about categories -- except for the things that I do know -- I'm wondering if and how such property could be expressed and proven in terms of universal properties, without actually manoevring inside the objects. For instance one may attempt to create a morphism $G\to H$ somehow appealing to the universal property of $\oplus$, and subsequently show this morphism is an isomorphism by chasing diagrams. But it seems likely that the existence of a structure theorem of some sort will be required.  This question may be considered trivial or weird for someone who's fluent with categories, I don't know.  contains quite a few references. I haven't really worked through any of them (yet) but I couldn't find anything helpful at first sight.  -- edit, So a more precise title would have been "Under what conditions that can be expressed in a universal way does $G\oplus G\cong H\oplus H$ imply $G\cong H$ ?" but I don't like long titles.  -- edit2, By the comment by Alexei Averchenko, maybe It's more natural to ask this question with the product instead of the coproduct. An answer to my question with the 'real' product would be appreciated too, obviously.
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We know that if a group is Abelian, then all its subgroups are normal. Also, if a group is nonabelian, it can contain a subgroup which is Abelian. Eg: The Dihedral group of order 2n, $D_{2n}$ is nonabelian for $n\geq 3$ but has $C_n$ sitting inside it. Now, my question is: "If a finite group is nonabelian, can it have a normal subgroup which is nonabelian?" In other words, can a normal subgroup of a nonabelian finite group be nonabelian? Thanks.
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If $G$ is an abelian group, every subgroup $H$ of $G$ is normal.  I searched for non-abelian finite groups $G$ , such that every subgroup is normal and GAP showed only the groups $G'\times Q_8$ , where $Q_8$ is the quaternion-group of order $8$ and $G'$ is an abelian group with the additional property that $C_4$ is not a subgroup of $G'$. The largest group I found with GAP was $C_{15}\times Q_8$.     Is it true that every non-abelian finite group $G$ with the property that every subgroup of $G$ is normal, is isomorphic to $G'\times Q_8$ with an abelian    group $G'$ without $C_4$ as a subgroup ?
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Are there four non-trivial groups $G_1,G_2,H_1,H_2$ pairwise non-isomorphic but such that $G_1 \times G_2$ is isomorphic to $H_1 \times H_2$?  My idea was to consider the abelian groups and something related to their decomposition as direct product of cyclic groups, but it doesn't work.