Question

1. Build up the operation tables for group G with orders 1, 2, 3 and 4 using the elements a, b, c, and e as the identity element in an appropriate way.
2. i. State the Lagrange’s theorem of group theory.
ii. For a subgroup H of a group G, prove the Lagrange’s theorem.
iii. Discuss whether a group H with order 6 can be a subgroup of a group with order 13 or not. Clearly state the reasons.

Answer 1

1.For the order 1 we obtain, that |G| = 1. Therefore G = {e}.

For the order 2 we obtain, that |G| = 2. Therefore G= {e,a}, where a² = 1. The respective table is:

$e \cdot e = e$

$e\cdot a = a$

$a \cdot e = a$

$a \cdot a = e$

For the order 3 we obtain, that |G| = 3. Therefore G={e,a,b}, where a³ = b³ = e and a^-1 = b. The respective table is:

$e \cdot e = e$ ; $a \cdot e = a$ ; $b \cdot e = b$

$e \cdot a = a$ ; $a \cdot a = a^2 = a^3 \cdot a^{-1} = e \cdot b = b$ ; $b\cdot a = a^{-1} \cdot a = e$

$e\cdot b = b$ ; $a \cdot b = a \cdot a^{-1} = e$ ; $b \cdot b = b^2 = a^{-2} = a^{-2} \cdot a^3 = a$

For the order 4 we obtain, that |G| = 4. Therefore G = {e,a,b,c}. Since there are 2 non-isomorphic groups of the order 4, we will consider the cyclic one. This means, that:

$e \cdot e = e$ ; $a \cdot e = a$ ; $b \cdot e = b$ ; $c \cdot e = c$

$e \cdot a = a$ ; $a \cdot a = a^2 = b$ ; $b \cdot a = a^2 \cdot a = a^3 = c$ ; $c \cdot a = a^3 \cdot a = e$

$e \cdot b = b$ ; $a \cdot b = a \cdot a^2 = a^3 = c$ ; $b \cdot b = b^2 = a^4 = e$ ; $c \cdot b = a^3 \cdot a^2 = a$

$e \cdot c = c$ ; $a \cdot c = a \cdot a^3 = a^4 = e$ ; $b \cdot c = a^2 \cdot a^3 = a^5 = a^1 = a$ ; $c \cdot c = a^3 \cdot a^3 = a^2 = b$

2.1)Let's state the Lagrange's theorem:

Let H be a subgroup of group G. Then |G| = [G:H] $\cdot$ |H|, where [G:H] is the number of right cosets of H in G.

2) For the proof we will consider the right cosets of H in G. Indeed, for any aH and bH either aH$\cap$ bH = $\emptyset$ or aH = bH. this statement is true due to the right cosets are the equivalent classes, and since if a,b belong to the same class then a^-1b belong ot H. So, if c belong to aH and bH, then a^-1c and c^-1b belong to H. Thus $a^{-1} \cdot c \cdot c^{-1} \cdot b = a^{-1} \cdot b \in H$ . So aH = bH. Now, since |aH| = |H| for every a from G(since f : H -> aH such that f(x) = ax is clearly bijective), we obtain that |G| = [G:H] $\cdot$ |H|, where [G:H] is the number of right cosets of H in G.

3) It is not true. As an obvious consequence of this theorem, we obtain that if |H| = 6, |G| $\neq$ 13, since 6 doesn't divide 13.