Cosets and Lagrange's Theorem

In $\mathbb Z_{12}$, let $H = \langle 3\rangle = \{0, 3, 6, 9\}$, a subgroup of order $4$ (Topic 3). The left cosets $g + H$: $0+H = \{0,3,6,9\} = H$, $1+H = \{1,4,7,10\}$, $2+H = \{2,5,8,11\}$. Trying $g=3$ gives nothing new: $3+H = \{3,6,9,0\} = H$ again. These three cosets are disjoint and their union is all $12$ elements of $\mathbb Z_{12}$, exactly as the partition claim predicts.

In $S_3$ (Topic 4) let $H = \{e, (12)\}$, order $2$. Compute the coset of $g=(123)$ both ways. Left: $(123)H = \{(123)e,\ (123)(12)\} = \{(123), (13)\}$. Right: $H(123) = \{e(123),\ (12)(123)\} = \{(123), (23)\}$. The two cosets share $(123)$ but are not the same set: $(13) \ne (23)$, so $gH \ne Hg$ for this $g$.

Let $H = A_3 = \{e, (123), (132)\}$ in $S_3$, order $3$. Since $|S_3|=6$ there can be at most $6/3=2$ cosets. $H$ itself is one. Pick any $g \notin H$, say $g=(12)$: $gH = \{(12)e, (12)(123), (12)(132)\} = \{(12), (23), (13)\}$, the three transpositions — accounting for all $6$ elements of $S_3$. So $S_3$ splits into exactly two cosets: the even permutations and the odd ones, and any other choice of $g \notin H$ gives this same second coset.

In $D_4$, the symmetries of a square, $|D_4| = 8$ (Topic 2). Let $H = \langle s\rangle$ for one reflection $s$; since $s^2=e$ and $s \ne e$, $|H|=2$. By Lagrange, $[D_4:H] = 8/2 = 4$: the reflection's cosets split the $8$ symmetries into exactly $4$ pairs.

Could a group of order $10$ have a subgroup of order $4$? Lagrange demands $4 \mid 10$, and it does not, since $10/4$ is not an integer — so the answer is no, with no search of the group required. The same check rules out a subgroup of order $3$, while one of order $5$ is at least not ruled out.

What are the possible orders of a subgroup of a group of order $30$? By Lagrange, only divisors of $30$ are possible: $1, 2, 3, 5, 6, 10, 15, 30$. Lagrange does not promise that all eight actually occur — only that nothing else can.

Take $g=(123)(45) \in S_5$, where $|S_5| = 120$ (Topic 4). Its cycle type is $3+2$, so $\operatorname{ord}(g) = \operatorname{lcm}(3,2) = 6$. Indeed $6 \mid 120$, exactly as corollary (1) demands.

$\mathbb Z_{12}$ is cyclic, so — unlike a general group — it has exactly one subgroup for each divisor of $12$. The divisors are $1,2,3,4,6,12$, giving the subgroups $\{0\}$, $\langle 6\rangle=\{0,6\}$, $\langle 4\rangle=\{0,4,8\}$, $\langle 3\rangle=\{0,3,6,9\}$, $\langle 2\rangle=\{0,2,4,6,8,10\}$, and $\mathbb Z_{12}$ itself: $6$ subgroups in all.

$|U(35)| = \varphi(35) = 24$, so a generator would need order $24$. Lagrange rules this out without computing a single order. Take any $a$ coprime to $35$. Modulo $5$, $a$ lies in $U(5)$, a group of order $4$, so corollary (2) gives $a^4 \equiv 1$ and hence $a^{12} = (a^4)^3 \equiv 1 \pmod 5$. Modulo $7$, $a$ lies in $U(7)$, of order $6$, so $a^6 \equiv 1$ and $a^{12} \equiv 1 \pmod 7$. Both $5$ and $7$ divide $a^{12}-1$, so $35$ does: $a^{12} \equiv 1 \pmod{35}$ for every $a \in U(35)$. Every element's order therefore divides $12$ and never reaches $24$, so $U(35)$ is not cyclic. Check with $a=2$: $2^{12} = 4096 = 35\cdot117+1$.

$\varphi(15)=|U(15)|=8$. Take $a=7$, coprime to $15$: Euler's theorem predicts $7^8 \equiv 1 \pmod{15}$. Direct computation confirms $7^8 = 5764801 = 15\cdot384320+1$.

Take $p=13$, $a=5$. Fermat predicts $5^{12}\equiv1\pmod{13}$. Checking: $5^{12}=244140625=13\cdot18780048+1$.

Find $7^{100} \bmod 13$. Since $13$ is prime and $13 \nmid 7$, $\operatorname{ord}(7)$ divides $12$, so $7^{12}\equiv1\pmod{13}$ and only the exponent mod $12$ matters: $100 = 8\cdot12+4$, so $7^{100}\equiv7^4\pmod{13}$. Now $7^2=49\equiv10$, so $7^4\equiv10^2=100\equiv9\pmod{13}$.