Question

state lagrange's theorem

Answer 1

This can be shown using the concept of left cosets of H in G. The left cosets are the equivalence classes of a certain equivalence relation on G and therefore form a partition of G. Specifically, x and y in G are related if and only if there exists h in H such that x = yh. If we can show that all cosets of H have the same number of elements, then each coset of H has precisely |H| elements. We are then done since the order of H times the number of cosets is equal to the number of elements in G, thereby proving that the order of H divides the order of G. Now, if aH and bH are two left cosets of H, we can define a map f : aH → bH by setting f(x) = ba−1x. This map is bijective because its inverse is given by {\displaystyle f^{-1}(y)=ab^{-1}y{\mbox{.}}}

This proof also shows that the quotient of the orders |G| / |H| is equal to the index [G : H] (the number of left cosets of H in G). If we allow G and H to be infinite, and write this statement as

{\displaystyle \left|G\right|=\left[G:H\right]\cdot \left|H\right|{\mbox{,}}}
then, seen as a statement about cardinal numbers, it is equivalent to the axiom of choice.

Answer 2

Answer:

$\huge\underline\mathrm\pink{Answer (◍•ᴗ•◍)}$

ʟᴀɢʀᴀɴɢᴇ ᴛʜᴇᴏʀᴇᴍ ɪs ᴏɴᴇ ᴏғ ᴛʜᴇ ᴄᴇɴᴛʀᴀʟ ᴛʜᴇᴏʀᴇᴍs ᴏғ ᴀʙsᴛʀᴀᴄᴛ ᴀʟɢᴇʙʀᴀ. ɪᴛ sᴛᴀᴛᴇs ᴛʜᴀᴛ ɪɴ ɢʀᴏᴜᴘ ᴛʜᴇᴏʀʏ, ғᴏʀ ᴀɴʏ ғɪɴɪᴛᴇ ɢʀᴏᴜᴘ sᴀʏ ɢ, ᴛʜᴇ ᴏʀᴅᴇʀ ᴏғ sᴜʙɢʀᴏᴜᴘ ʜ ᴏғ ɢʀᴏᴜᴘ ɢ ᴅɪᴠɪᴅᴇs ᴛʜᴇ ᴏʀᴅᴇʀ ᴏғ ɢ. ᴛʜᴇ ᴏʀᴅᴇʀ ᴏғ ᴛʜᴇ ɢʀᴏᴜᴘ ʀᴇᴘʀᴇsᴇɴᴛs ᴛʜᴇ ɴᴜᴍʙᴇʀ ᴏғ ᴇʟᴇᴍᴇɴᴛs. ᴛʜɪs ᴛʜᴇᴏʀᴇᴍ ᴡᴀs ɢɪᴠᴇɴ ʙʏ ᴊᴏsᴇᴘʜ-ʟᴏᴜɪs ʟᴀɢʀᴀɴɢᴇ. ɪɴ ᴛʜɪs ᴀʀᴛɪᴄʟᴇ, ʟᴇᴛ ᴜs ᴅɪsᴄᴜss ᴛʜᴇ sᴛᴀᴛᴇᴍᴇɴᴛ ᴀɴᴅ ᴘʀᴏᴏғ ᴏғ ʟᴀɢʀᴀɴɢᴇ ᴛʜᴇᴏʀᴇᴍ ɪɴ ɢʀᴏᴜᴘ ᴛʜᴇᴏʀʏ, ᴀɴᴅ ᴀʟsᴏ ʟᴇᴛ ᴜs ʜᴀᴠᴇ ᴀ ʟᴏᴏᴋ ᴀᴛ ᴛʜᴇ ᴛʜʀᴇᴇ ʟᴇᴍᴍᴀs ᴜsᴇᴅ ᴛᴏ ᴘʀᴏᴠᴇ ᴛʜɪs ᴛʜᴇᴏʀᴇᴍ ᴡɪᴛʜ ᴛʜᴇ ᴇxᴀᴍᴘʟᴇs. sᴇᴛ ᴛʜᴇᴏʀʏ sᴇᴛs ᴍᴀᴛʜs ғɪɴɪᴛᴇ ᴀɴᴅ ɪɴғɪɴɪᴛᴇ sᴇᴛs sᴜʙsᴇᴛ ᴀɴᴅ sᴜᴘᴇʀsᴇᴛ ʟᴀɢʀᴀɴɢᴇ ᴛʜᴇᴏʀᴇᴍ sᴛᴀᴛᴇᴍᴇɴᴛ ᴀs ᴘᴇʀ ᴛʜᴇ sᴛᴀᴛᴇᴍᴇɴᴛ, ᴛʜᴇ ᴏʀᴅᴇʀ ᴏғ ᴛʜᴇ sᴜʙɢʀᴏᴜᴘ ʜ ᴅɪᴠɪᴅᴇs ᴛʜᴇ ᴏʀᴅᴇʀ ᴏғ ᴛʜᴇ ɢʀᴏᴜᴘ ɢ. ᴛʜɪs ᴄᴀɴ ʙᴇ ʀᴇᴘʀᴇsᴇɴᴛᴇᴅ ᴀs; |ɢ| = |ʜ| ʙᴇғᴏʀᴇ ᴘʀᴏᴠɪɴɢ ᴛʜᴇ ʟᴀɢʀᴀɴɢᴇ ᴛʜᴇᴏʀᴇᴍ, ʟᴇᴛ ᴜs ᴅɪsᴄᴜss ᴛʜᴇ ɪᴍᴘᴏʀᴛᴀɴᴛ ᴛᴇʀᴍɪɴᴏʟᴏɢɪᴇs ᴀɴᴅ ᴛʜʀᴇᴇ ʟᴇᴍᴍᴀs ᴛʜᴀᴛ ʜᴇʟᴘ ᴛᴏ ᴘʀᴏᴠᴇ ᴛʜɪs ᴛʜᴇᴏʀᴇᴍ. ᴡʜᴀᴛ ɪs ᴄᴏsᴇᴛ? ɪɴ ɢʀᴏᴜᴘ ᴛʜᴇᴏʀʏ, ɪғ ɢ ɪs ᴀ ғɪɴɪᴛᴇ ɢʀᴏᴜᴘ, ᴀɴᴅ ʜ ɪs ᴀ sᴜʙɢʀᴏᴜᴘ ᴏғ ɢ, ᴀɴᴅ ɪғ ɢ ɪs ᴀɴ ᴇʟᴇᴍᴇɴᴛ ᴏғ ɢ, ᴛʜᴇɴ; ɢʜ = { ɢʜ: ʜ ᴀɴ ᴇʟᴇᴍᴇɴᴛ ᴏғ ʜ } ɪs ᴛʜᴇ ʟᴇғᴛ ᴄᴏsᴇᴛ ᴏғ ʜ ɪɴ ɢ ᴡɪᴛʜ ʀᴇsᴘᴇᴄᴛ ᴛᴏ ᴛʜᴇ ᴇʟᴇᴍᴇɴᴛ ᴏғ ɢ ᴀɴᴅ ʜɢ = { ʜɢ: ʜ ᴀɴ ᴇʟᴇᴍᴇɴᴛ ᴏғ ʜ } ɪs ᴛʜᴇ ʀɪɢʜᴛ ᴄᴏsᴇᴛ ᴏғ ʜ ɪɴ ɢ ᴡɪᴛʜ ʀᴇsᴘᴇᴄᴛ ᴛᴏ ᴛʜᴇ ᴇʟᴇᴍᴇɴᴛ ᴏғ ɢ. ɴᴏᴡ, ʟᴇᴛ ᴜs ʜᴀᴠᴇ ᴀ ᴅɪsᴄᴜssɪᴏɴ ᴀʙᴏᴜᴛ ᴛʜᴇ ʟᴇᴍᴍᴀs ᴛʜᴀᴛ ʜᴇʟᴘs ᴛᴏ ᴘʀᴏᴠᴇ ᴛʜᴇ ʟᴀɢʀᴀɴɢᴇ ᴛʜᴇᴏʀᴇᴍ. ʟᴇᴍᴍᴀ : ɪғ ɢ ɪs ᴀ ɢʀᴏᴜᴘ ᴡɪᴛʜ sᴜʙɢʀᴏᴜᴘ ʜ, ᴛʜᴇɴ ᴛʜᴇʀᴇ ɪs ᴀ ᴏɴᴇ ᴛᴏ ᴏɴᴇ ᴄᴏʀʀᴇsᴘᴏɴᴅᴇɴᴄᴇ ʙᴇᴛᴡᴇᴇɴ ʜ ᴀɴᴅ ᴀɴʏ ᴄᴏsᴇᴛ ᴏғ ʜ. ʟᴇᴍᴍᴀ : ɪғ ɢ ɪs ᴀ ɢʀᴏᴜᴘ ᴡɪᴛʜ sᴜʙɢʀᴏᴜᴘ ʜ, ᴛʜᴇɴ ᴛʜᴇ ʟᴇғᴛ ᴄᴏsᴇᴛ ʀᴇʟᴀᴛɪᴏɴ, ɢ ∼ ɢ ɪғ ᴀɴᴅ ᴏɴʟʏ ɪғ ɢ ∗ ʜ = ɢ ∗ ʜ ɪs ᴀɴ ᴇǫᴜɪᴠᴀʟᴇɴᴄᴇ ʀᴇʟᴀᴛɪᴏɴ. ʟᴇᴍᴍᴀ : ʟᴇᴛ s ʙᴇ ᴀ sᴇᴛ ᴀɴᴅ ∼ ʙᴇ ᴀɴ ᴇǫᴜɪᴠᴀʟᴇɴᴄᴇ ʀᴇʟᴀᴛɪᴏɴ ᴏɴ s. ɪғ ᴀ ᴀɴᴅ ʙ ᴀʀᴇ ᴛᴡᴏ ᴇǫᴜɪᴠᴀʟᴇɴᴄᴇ ᴄʟᴀssᴇs ᴡɪᴛʜ ᴀ ∩ ʙ = ∅, ᴛʜᴇɴ ᴀ = ʙ.

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