Can someone give me an example of a group which has exactly $2$ proper, non-trivial subgroups?  I appreciate that there is no context with this question but there's no really much else I can say.
122933
Suppose $G$ is a group and has exactly two nontrivial proper subgroups. Prove that $G$ is cyclic and $|G|=pq$ where p,q are distinct primes or $G$ is cyclic and $|G|=p^3$ where $p$ is a prime. Usually my tactic to show a group is cyclic is find an elements such that its order and order of the group are the same. But here I cannot do so because I have no informatio regarding order of elements. So I stuck. Can anyone help me?
122713
What is a (simple) example of a group $G$ with an endomorphism $G\rightarrow G$ That is surjective, but not injective?
73465
I have been going through some material on elementary group theory () and in one place it states:  A subgroup which contains half of all elements, i.e. for which $|G|=2|H|$, is a normal subgroup.  Where a subgroup $H\subset G$ is called a normal subgroup if $gHg^{-1}=H~\forall g\in G$. How would I go about proving this? Sorry if the answer exists somewhere, but I tried searching and could not find it.
20677
In , people discussed that there cannot be cyclics subgroups of order 9 in $S_6$.   How does one reason that the subgroups isomorphic to $C_3 \times C_3$ (the other candidates to be subgroups of order 9) are precisely the ones given by two disjoint permutations $\langle (123),(456) \rangle$
61950
So if I'm given $H$, when I can conclude there is a group $G$ such that $H\cong G/Z(G)$? It's easy to show that non-trivial cyclic groups are not of this form. More generally, any group with the property that all finitely generated subgroups are cyclic (e.g. $\mathbb{Q},\mathbb{Q}/\mathbb{Z}$) cannot be of this form. Are there are any others?
52992
True or False. Every subgroup of a simple group is itself simple.  Our solution:The statement is false.  As a counter example, let $G$ be equal to the group of even permutation of $10$ letters and $H$ is the subgroup of $G$ which is $$H=\{e,(12)(34),(56)(78),(12)(34)(56)(78)\}.$$ $H$ is not simple since $K=\{e,(12)(34)\}$ is also normal in $H$.
32630
Well, this is an exercise problem from Herstein which sounds difficult:   How does one prove that if $|G|&gt;2$, then $G$ has non-trivial automorphism?   The only thing I know which connects a group with its automorphism is the theorem, $$G/Z(G) \cong \mathcal{I}(G)$$ where $\mathcal{I}(G)$ denotes the Inner- Automorphism group of $G$. So for a group with $Z(G)=(e)$, we can conclude that it has a non-trivial automorphism, but what about groups with center?
33599
If $G$ is a finite group of order $n,$ and the number of divisors of $n$ is $k,$ can $G$ have fewer than $k$ subgroups?  A cyclic group $G$ of order $n$ has exactly one subgroup for each divisor of $n$, so in this case $G$ has exactly $k$ subgroups.  For the alternating group $G=A_4,$ $n=12$ and $k=6.$ There is no subgroup of order $6$ in $A_4,$ yet there are ten subgroups, which exceeds $k=6$ for this case. I began to wonder if there could be a group for which so many divisors of the order of $G$ had no subgroups of that order, that there wound up being fewer subgroups in $G$ than the number of divisors of $|G|,$ hence my question.
101145
What is the relationship between abelian subgroup of a  non-abelian group(when exist, example, theorem)?? any thing such link regarding the question would help.   I apologize for previous post-error.
34133
i) Does there exist a group $G$ with a normal subgroup $H$ such that $H , G/H$ are abelian but $G$ is not ?   ii) Does there exist a group $G$ with a normal subgroup $H$ such that $H , G/H$ are cyclic but $G$ is not ?
5791
I'm looking for verification:  My claim: If $G$ is a finite group and $H$ is a (proper)subgroup of index $k&gt;1$, where $k! \leq |G|$, then $G$ is not simple.  Proof: Consider the set of left cosets of $G$ by $H$:  $$G/H = \{H,g_1H,\dots,g_kH\}$$  Then $\phi: G \rightarrow S_{G/H}$ is a homomorphism where we define:  $$\phi(g)[g_iH] = g g_iH$$  ($S_{G/H}$ is the group of permutations on elements of $G/H$ and of course $S_{G/H} \cong S_k$.)  Also $\ker \phi \neq G$ because $\phi(g_1)$ is clearly not the identity in $S_{G/H}$.  Because $\phi$ is a homomorphism,  $$ G/\ker\phi \cong \text{im}\phi$$  Then $|\text{im} \phi| \leq |S_k| = k!$ and if $k! \leq |G|$ then $|\ker\phi| &gt; 1$ and $\ker\phi$ is a non-trivial normal subgroup of $G$.  Example: If a group of order $36$ has a subgroup of index $3$ or $4$ then the group is not simple.
12667
So I'm having trouble with this problem. I know that the definition of a simple group means that the group has no nontrivial subgroups. I know that this can be proven somehow with the help of the converse of Lagrange's Theorem for Abelian groups: If G is abelian of order n, and d is a divisor of n, then G has a subgroup of order d.   My attempt: (=>)Assume that G is a finite abelian group and is simple, then G has no nontrivial normal  subgroups.  (Now I don't know how to show that this implies that G has order p, where p is prime.  (&lt;=)Assume that G is a finite abelian group with order p, where p is a prime. (Since the order of p is prime then what does this mean?)  Edit: Can someone check my new attempt at the proof?   (=>) Suppose G is a simple finite abelian group. Suppose for the sake of contradiction that G does not have prime order, then |G|=p*k where p is a prime number and k is an integer such that k>1. Then G has an element of order p. Let the element of order p be called x. Then , the subgroup generated by x, is of order p and  is not all of G. Since G is abelian, this subgroup is normal, which leads us to a contradiction. Therefore, G must have prime order.   (&lt;=) Suppose that G is a finite abelian group and it’s order is p, a prime. Since G has prime order, then the only two subgroups of G are the trivial subgroup and the group G. Then, by definition the group G is simple since there are no nontrivial proper subgroups, and thus no nontrivial normal subgroups.
69096
Let $G$ be a group of order 24, and we shall assume there there exist a non-normal 2-sylow group in $G$. i want to show that it's not simple.  first i have showed that there are exactly three 2-sylow group's.   and now i am trying to show that there is a non trivial homomorphism from $G$ to $S_3$. i know that $S_3\cong D_3$  and indeed there are three 2-sylow groups in $D_3$ , the reflection's. and there is one $3-$sylow group in $D_3$  so i wanted to creat a homomorphism that take the different p-sylow groups in $G$  and set them in right $p-$sylow group's in $D_3$  but for that i need to show that the intersection of every two $2-$sylow gouts in $G$  is trivial. and i can't manage to do that....
11401
Suppose $G$ is a finite group and I know for every $k \leq |G|$ that exactly $n_k$ elements in $G$ have order $k$. Do I know what the group is? Is there a counterexample where two groups $G$ and $H$ have the same number of elements for each order, but $G$ is not isomorphic to $H$? I suspect that there is, but I haven't thought of one.
90448
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.
118880
Problem: Let $G$ be a group of order $108 = 2^23^3$. Prove that $G$ has a proper normal subgroup of order $n \geq 6$.  My attempt: From the Sylow theorems, if $n_3$ and $n_2$ denote the number of subgroups of order $27$ and $4$, respectively, in $G$, then $n_3 = 1$ or $4$, since $n_3 \equiv 1$ (mod $3$) and $n_3~|~2^2$, and $n_2 = 1, 3, 9$ or $27$, because $n_2~|~3^3$.  Now, I don't know what else to do. I tried assuming $n_3 = 4$ and seeing if this leads to a contradiction, but I'm not even sure that this can't happen. I'm allowed to use only the basic results of group theory (the Sylow theorems being the most sophisticated tools).  Any ideas are welcome; thanks!
93320
Let $G$ be a characteristically simple finite group, i.e. it has no nontrivial characteristic subgroups. Prove there is some simple group $T$ such that $G \cong T \times T \times \cdots \times T$.   No idea how to start this one. I have tried to induct on the size of $G$ to no avail. Any help would be appreciated!