I am not able to prove this, could any one help me?     $G$ be a finite group, $G$ has exactly one proper subgroup. We need to prove that $G$ is cyclic and $|G|=p^2$ where $p$ is prime.   Thank you.
122412
Suppose that $G$ is a group that has exactly one nontrivial proper subgroup. Then we have to show that $G$ is cyclic and order of $G$ is $p^2$ where $p$ is prime.   I tried as, if $a$ and $b$ two element of $G$ of order $p$ and $q$ then $\langle a \rangle$ and $\langle b \rangle$ are two proper subgroups – a contradiction. Thus every element of $G$ have order some power of $p$. After this I am stuck. Thanks for help.
93271
Let $G$ be a finite abelian group s.t. it contains a subgroup $H_{0} \neq (e)$ which lies in every subgroup $H \neq (e) $. Prove that $G$ must be cyclic.   Also what can be said about $o(G)$ ?    I'm clueless about this problem from Herstein. I tried to think $G$ might be a solvable group and its subgroups are all normal (as $G$ is abelian) . But I'm unable to get further.
2653
Problem Suppose G is a finite group of order $n$ which has a unique subgroup of order $d$ for each $d\mid n$. Prove that $G$ must be a cyclic group.  My idea:   I try to prove it by induction. Let $p|n$ be a prime. Then by condition, there exists a unique subgroup $H$ of order $n/p$. Since $|gHg^{-1}|=|H|$, we must have $gHg^{-1}=H$ by the uniqueness part of the condition. So, H is a normal subgroup. Now, $|G/H|=p$ and thus $G/H=\langle x\rangle$ where $x^p \in H$.    However, I cannot continue.  My another idea is that first consider the case when $|G|$ is a power of some prime $p$. But, it still doesn't work.
69131
I'm considering finite groups only. Cayley's theorem says the a group $G$ is isomorphic to a subgroup of $S_{|G|}$. I think it's interesting to ask for smaller values of $n$ for which $G$ is a subgroup of $S_n$. Obviously, it's not always possible to do better than Cayley's theorem. But sometimes it is possible (for example, $\mathbb{Z}_6$ as a subgroup of $S_5$).  So I'm asking:        Given a finite group $G$, is there an algorithmic way to find or approximate the minimal $n$ for which $G$ is isomorphic to a subgroup of $S_n$?   If the answer to $(1)$ is not known, is it known for specific classes of groups?   In particular, for finite abelian groups, is it true that for a prime $p$, the minimal $n$ for $\mathbb{Z}_{p^{t_1}} \times \mathbb{Z}_{p^{t_2}}$ is $p^{t_1}+p^{t_2}$ (I can prove that is is true for different primes $p_1$ and $p_2$, but have problems when it's the same prime in both factors).      Thanks!
18013
I am participating in an Introductory course in groups and I have the following question:     Let $G$ be a finite group of order $8$. Prove that $G$ has a subgroup of order $4$ and a subgroup of order $2$.   i know how to prove that $G$ has a subgroup of order $2$ but i can't show the existence of a subgroup of order 4. I am only a begginer in groups so please take this into account.
104048
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?
80770
Just wanted some feedback to ensure I did not make any mistakes with this proof. Thanks!  Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$, and $|G| &gt; 1$, so for some $x\in G$ we have $x\neq 1$ and $\langle x\rangle\leq G$, so $\langle x\rangle = G$. Suppose $x$ has either infinite order or even order. Then $\langle x^2\rangle \leq G$ but $\langle x^2\rangle \neq \langle x \rangle$, a contradiction. So $x$, and therefore $G$, has odd order, and by Feit-Thompson, $G\cong Z_p$ for some prime $p$.  Edit: Thanks, I see that Feit-Thompson is too much.  Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$, and $|G| &gt; 1$, so for some $x\in G$ we have $x\neq 1$ and $\langle x\rangle\leq G$, so $\langle x\rangle = G$. Suppose $x$ has infinite order. Then $\langle x^2\rangle \leq G$ but $\langle x^2\rangle \neq \langle x \rangle$, a contradiction. So $x$, and therefore $G$, has finite order. Suppose $x$ has composite order $n$ so for some $p &gt; 1$ that divides $n$, $\langle x^p \rangle$ is a proper non-trivial subgroup of $G$, so $G$ is not simple. So $G$ is a cyclic group of prime order.
79950
$G$ is a Group of order $pq$, if $G$ has exactly one subgroup of order $p$ and another with order $q$, then $G$ is cyclic ($p,q$ are prime)  Here's what I have so far, I'm not sure if it is correct.  Let $H, K$ be subgroups of $G$ $|H| = p$  $|K| = q$   We know that by Lagrange's Theorem:  $\forall a \in H : |a|=p$ where $a \neq e$ $\Rightarrow H = &lt;a&gt;$  Similarly: $\forall a \in K : K = &lt;a&gt;$  We also know that: $\forall a \in G : |a| | pq \Rightarrow |a| = p$ or $|a| = q$   $|a| = p \Rightarrow a \in H$ since there is exactly one subgroup of order $p$  $|a| = q \Rightarrow a \in K$ for similar reasons  This means every element in $G$ is either in $H$ or $K$, which are both cyclic. That means $G$ is cyclic.  What am I missing here? Have I made any mistakes and/or is the proof incomplete?
34406
$A=\{z\in\mathbb{C}: z^{p^n}=1\;\mathrm{for\;some\;integer}\;n\geq1\}$. I have to prove:  1) Every proper subgroup of $A$ is cyclic,  2) if $B,C$ are subgroups of $A$ then $B\subseteq C$ or $C\subseteq B$,  3) for every $n\geq0$ there exists an unique subgroup of $A$ with $p^n$ elements.  Could you help me please?
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!
66181
Suppose we have a finite group $G$ of finite order $n$. For every $d\mid n$, $G$ has at most one subgroup of order $d$. Show that $G$ is cyclic.
47472
Prove that if $|G| = p^n$ then $G$ has a subgroup of order $p^m$ for all $0\le m &lt;n.$   Since $G$ is of prime-power order I know $|Z(G)| \ne e$ so there is an $a\in Z(G)$ with order $p$ such that $p \mid |Z(G)|$. Now, the subgroup generated by is normal since it's a subgroup of the center. How can I get this normal subgroup to be less than $n$?
47742
Let G be a finite group and H be a proper subgroup. Prove that the union of the conjugates of H is not the whole of G.  Thanks for any help
29567
Suppose $G$ is finite and $G=H\cup K\cup L$ for proper subgroups $H,K,L$. Show that $|G:H|=|G:K|=|G:L|=2$.  What I did: so if some of $H,K,L$ is contained in another, then we have $G$ being a union of two proper subgroups, which is impossible due to another result. So none of $H,K,L$ is contained is each other. If some element $h$ belongs to only $H$ and $k$ belongs to only $K$, then its product $hk$ cannot be in $H$ or $K$, so must belong to only $L$.
81331
Let G be a group with |G| = pq. (p and q are both prime).  How can we prove that every proper subgroup of G is cyclic?
78108
I need some help on this one.     $G$ is a group. If $N$ is a subgroup of $G$ contained in the centre of $G$ and $G/N$ is cyclic, show that $G$ is Abelian.   My attempt is only half way and stuck at the bit at the end...     If $G$ can be shown to be cyclic, then it is Abelian, since $G=&lt;g&gt;$ then $g_1,g_2 \in G$ are $g_1=g^n,g_2=g^m$ for some $n,m$ and $g_1g_2=g^ng^m=g^{n+m}=g^{m+n}=g_2g_1$.       By definition of a centre of $G$, $\forall n \in N$, $gn=ng$ for any element in $G$. Namely, $N$ must be a normal subgroup.   (Here, I pondered if I should just proceed or change gears to looking at homomorphisms of $G$ since $N$ is normal).      Now $G/N$ is cyclic so $G/N=&lt;gN&gt;$ for some $g \in G$. Then, $\forall g'N \in G/N$ there is some $m$ so that $(gN)^m=g^mN=g'N$. Therefore, for the representative elements of $G/N$, I have $g^m$.   Here I get stuck. So if $g'N$ then I have that each $g'$ is cyclic but as an equivalent class, I cannot guarantee that all $g'$ can be represented in the form $g^m$. The only thing I can guess is the canonical map $p:G \to G/N$ that might be relevant to solving this but again, I don't know how specifically it would help.  Does anyone know how to solve this?
7205
$P$ and $Q$ are subgroups of a group $G$. How can we prove that $P\cap Q$ is a subgroup of $G$? Is $P \cup Q$ a subgroup of $G$?   Reference: Fraleigh p. 59 Question 5.54 in A First Course in Abstract Algebra.