Show that $H$ is unique and $G$ is isomorphic to $\mathbb{Z}_{77}$ Let $G$ be a group of order $77$.

*

*Show that $G$ has a subgroup $H$ of order $11$.

*Show that $H$ is unique, and hence normal.

*Conclude that $G$ is isomorphic to $\mathbb{Z}_{77}$
For $(1)$, the answer is simple, it is enough to assume that there are only elements other than the identity with order $7$, in this way we arrive at a contradiction.
My problem is in $(2)$, how can I show that $H$ is indeed unique?
Once this is done, it is easy to prove that it is normal since if $a\in G$, then $a^{-1}Ha$ is a subgroup of order $11$, so $a^{-1}Ha=H$.
For $(3)$, I think it is sufficient to show that the subgroup of order $7$ is also unique and therefore cyclic and isomorphic to $\mathbb{Z}_{77}$.
I haven't seen sylow's theorems yet. Thanks in advance.
 A: To prove (2) you can proceed as follows, without an appeal to Sylow, only this one: in general, if $X,Y$ are subgroups of a finite group $G$, then the cardinality of the set $XY=\{xy: x\in X, y\in Y\}$ is
$$\#XY=\frac{|X|\cdot|Y|}{|X \cap Y|}.$$
Now you can use this to show that $H$ is normal. If not, then there is a $g \in G$ with $H \neq H^g$ (where $H^g=\{g^{-1}hg:h \in H\}$, which is a subgroup of $G$ and isomorphic to $H$). Since $|H|=11$, Lagrange's Theorem implies $H \cap H^g=1$. The formula above now yields that the set $HH^g$ has $121$ elements, which is of course absurd since $|G|=77$. So $H$ is normal. 
If $K$ is a subgroup of order $7$ (it exists by Cauchy's Theorem), as a bonus of the above formula and the fact that $H \cap K=1$ you get, $G=HK$.
Now proving $(3)$ requires somewhat more sophistication, namely normalizers and centralizers, in particular the so-called N/C Theorem and I hope this is covered in your book: if $X$ is a subgroup of a group $G$ then $N_G(X)/C_G(X)$ can be homomorphically embedded in $Aut(X)$, the group of automorphisms of $X$. Applying this to $H$ gives $N_G(H)/C_G(H)$ is isomorphic to a subgroup of $Aut(H) \cong C_{10}$. Since $H$ is normal, $N_G(H)=G$ and in addition, $|N_G(H)/C_G(H)|$ must divide $|G|=77$ and divides of course $|C_{10}|=10$. So, $G=C_G(H)$, that is $H \subseteq Z(G)$ and it follows that $K$ is also normal in $G=HK$. I leave it to you to infer that $G \cong C_7 \times C_{11} \cong C_{77}$.
A: This is an argument for 2) $\Rightarrow$ 3) (where the normality of the unique $H$ is not used, actually) by using the Class Equation. Now, $|G|=7\cdot11$. Let's assume $|Z(G)|=7$; therefore, the noncentral elements $g\in G$ have centralizer of order $7$, whence $\frac{|G|}{|C_G(g)|}=11$ for every noncentral $g\in G$, and finally (Class Equation) $7(11-1)=11k$, where $k$ is the number of noncentral conjugacy classes of $G$: contradiction. Thus, $|Z(G)|\ne 7$. The same argument works by swapping $11$ and $7$, whence  $|Z(G)|\ne 11$ as well. Therefore, either $|Z(G)|=7\cdot 11$ and $G$ is cyclic, or $Z(G)=\{e\}$. This latter is ruled out again by the Class Equation if $H\le G$ such that $|H|=11$ is unique; in fact, the Class Equation $77=1+7l_0+11l_1$ (where $l_0+l_1=k$) has the only solution $(l_0,l_1)=(3,5)$, in contradiction with the uniqueness of the subgroup of order $11$.
A: for  Question 1 and 2
if G is a Group of order p×q, where p and q are prime, p<q and p does not divide q-1, then G is cyclic. and Every cyclic group is abelian. If group is abelian then subgroup is also abelian. Every Subgroup of an Abelian Group is normal.

*

*Say you have an element a of order q, generating a subgroup ${H_q⊂G}$. Now by Lagrange's theorem ${H_q}$ is the unique subgroup of order q in G, hence it is normal. To see this, note that for any g∈G, the conjugate subgroup ${g H_qg^{−1}}$ also has order q, hence ${gH_qg^{−1}=Hq}$ for all g∈G. It follows that the quotient group ${G/H_q}$ is cyclic of order p. Choose a generator b of G/Hp and consider once of its preimages b′ in G. Note that b′ is not in ${H_q}$. Then the image of ${b′^p}$ will be in ${Hq}$, hence b′ has order either p or pq. If it is the latter, ${b′^q}$ has order p.

*The divisors of pq are 1, p, q, and pq. If a is an element of order pq, then ${a^p}$ is of order q and vice versa. So we may assume that you’ve taken an element g of G that’s of order p. It generates a subgroup of order only p, so there are other elements. If your next element is of order pq, you’re done, and if that next element is of order q, you’re also done.

For question 3. Below link could be helpful.
[1]: https://groupprops.subwiki.org/wiki/Classification_of_groups_of_order_a_product_of_two_distinct_primes
