# Show that $W = \langle v_1 + \dots + v_n\rangle$ is a submodule of $V$ .

Let $G = S_n$. Let $V$ be a vector space over a field $F$ with basis $\{v_1,\dots,v_n\}$. Then $V$ is an $FG$-module with action defined by setting $g · v_i = v_{g(i)}$ for all $g \in G$ and $1\leq i \leq n$ and extending linearly.

i) Show that $W = \langle v_1 +\dots+ v_n\rangle$ is a submodule of $V$.

I can show that for a finite basis that $W$ would be a submodule.

Let $G = S_3$ is the symmetric group on $3$ letters. Let $V$ be a vector space over a field $F$ with basis $B = \{v_1,v_2,v_3\}$. Define an action of $G$ on $V$ by setting $g·v_i =v_{g(i)}$ for $g\in G, i=1,2,3$ and extending linearly.

So, if $g = (1, 2, 3)$ then $$g·v_1 =v_2\\ g·v_2 =v_3\\ g·v_3 =v_1,$$ and if $$v=\sum_{i=1}^3 \lambda_iv_i \in V$$ then

$$g \cdot v = \sum_{i=1}^3 \lambda_ig \cdot v_i = \lambda_1v_2 + \lambda_2v_3 + \lambda_3v_1.$$ If $W=\langle v_1+v_2+v_3\rangle$, then $g·w=w$ for all $g \in G$. So $$W = \langle v_1 + v_2 + v_3\rangle$$ is called a submodule of $V$.

But I am unsure how to apply this to an infinite amount. Thanks.

• And what are the lovely squares supposed to represent? – 5xum Feb 26 '14 at 9:11
• sorry I can't see any squares, perhaps my use of lamda isn't compatible? – ZZS14 Feb 26 '14 at 9:16
• The symbol before v1 in your title and at the end of the formula for $W$. Also, put all formulas into mathjax (dollar signs at both ends). – 5xum Feb 26 '14 at 9:22
• is that any better? – ZZS14 Feb 26 '14 at 9:27

There is no "infinite" amount. Your proof for $3$ works just as well for a general $n$.
Take any $g\in S_n$ and any sum
$$v=\lambda \left(\sum_{i=1}^n v_i\right) \in W$$ Now $$g\cdot v=g\cdot \sum_{i=1}^n \lambda v_i = \sum_{i=1}^n \lambda v_{g(i)}=\lambda\sum_{i=1}^n v_{g(i)}=v.$$
• oh my confusion i suppose starts from how i would rewrite $$g·v_1 =v_2\\ g·v_2 =v_3\\ g·v_3 =v_1,$$ for n amount, like how would i 'decide' which $v_i$ goes to which – ZZS14 Feb 26 '14 at 9:41
• You don't know which $v_i$ goes where. You do know it goes to some other $v_j$, meaning that any sum of $v_i$ will still be a sum of $v_i$ after applying any $g$. – 5xum Feb 26 '14 at 10:03
• thanks, i understand now. would it be enough to say after that, that for some $w=v_1+...+v_n$ then g.w=w for all g in G so we can conclude that W=<\$v_1+....v_n> as required? – ZZS14 Feb 26 '14 at 10:19